Kuangbin专题七线段树

标签: 线段树

  没写完,还有五题,好像是乱七八糟建模乱七八糟维护什么的,在家里真的没心思写题目,先发博客吧以后再补换个代码量少的专题写写。

  

A - 敌兵布阵

 HDU - 1166

C国的死对头A国这段时间正在进行军事演习,所以C国间谍头子Derek和他手下Tidy又开始忙乎了。A国在海岸线沿直线布置了N个工兵营地,Derek和Tidy的任务就是要监视这些工兵营地的活动情况。由于采取了某种先进的监测手段,所以每个工兵营地的人数C国都掌握的一清二楚,每个工兵营地的人数都有可能发生变动,可能增加或减少若干人手,但这些都逃不过C国的监视。 
中央情报局要研究敌人究竟演习什么战术,所以Tidy要随时向Derek汇报某一段连续的工兵营地一共有多少人,例如Derek问:“Tidy,马上汇报第3个营地到第10个营地共有多少人!”Tidy就要马上开始计算这一段的总人数并汇报。但敌兵营地的人数经常变动,而Derek每次询问的段都不一样,所以Tidy不得不每次都一个一个营地的去数,很快就精疲力尽了,Derek对Tidy的计算速度越来越不满:"你个死肥仔,算得这么慢,我炒你鱿鱼!”Tidy想:“你自己来算算看,这可真是一项累人的工作!我恨不得你炒我鱿鱼呢!”无奈之下,Tidy只好打电话向计算机专家Windbreaker求救,Windbreaker说:“死肥仔,叫你平时做多点acm题和看多点算法书,现在尝到苦果了吧!”Tidy说:"我知错了。。。"但Windbreaker已经挂掉电话了。Tidy很苦恼,这么算他真的会崩溃的,聪明的读者,你能写个程序帮他完成这项工作吗?不过如果你的程序效率不够高的话,Tidy还是会受到Derek的责骂的. 

Input

第一行一个整数T,表示有T组数据。 
每组数据第一行一个正整数N(N<=50000),表示敌人有N个工兵营地,接下来有N个正整数,第i个正整数ai代表第i个工兵营地里开始时有ai个人(1<=ai<=50)。 
接下来每行有一条命令,命令有4种形式: 
(1) Add i j,i和j为正整数,表示第i个营地增加j个人(j不超过30) 
(2)Sub i j ,i和j为正整数,表示第i个营地减少j个人(j不超过30); 
(3)Query i j ,i和j为正整数,i<=j,表示询问第i到第j个营地的总人数; 
(4)End 表示结束,这条命令在每组数据最后出现; 
每组数据最多有40000条命令 

Output

对第i组数据,首先输出“Case i:”和回车, 
对于每个Query询问,输出一个整数并回车,表示询问的段中的总人数,这个数保持在int以内。 

Sample Input

1
10
1 2 3 4 5 6 7 8 9 10
Query 1 3
Add 3 6
Query 2 7
Sub 10 2
Add 6 3
Query 3 10
End 

Sample Output

Case 1:
6
33
59

单点更新,区间维护和值。


#include <queue>
#include <cstdio>
#include <cstring>
#include <utility>
#include <iostream>
#include <map>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
using namespace std;
typedef pair<int, int> P;
typedef long long ll;
#define N 200010
#define M 3645
const int INF = 0x3f3f3f3f;
const double eps = 1e-5;
const double PI = acos(-1);
#define fi first
#define se second
#define L o<<1
#define R o<<1|1
#define tl tree[o].l
#define tr tree[o].r
#define tw tree[o].w
#define rep(i, lll, nnn) for(int i = (lll); i <= (nnn); i++)

struct Node {
    int l, r, w;
}tree[N];

void build(int o, int l, int r)
{
    tl = l; tr = r;
    if(l == r) {
        scanf("%d", &tw);
        return;
    }
    int mid = (l + r) >> 1;
    build(L, l, mid);
    build(R, 1 + mid, r);
    tw = tree[L].w + tree[R].w;
}

void add(int o, int i, int w)
{
    if(tl == tr) {
        tw += w;
        return;
    }
    tw += w;
    int mid = (tl + tr) >> 1;
    if(i <= mid) add(L, i, w);
    else add(R, i, w);
}

int query(int o, int l, int r)
{
    //cout << l << ' ' << r<<endl;
    if(tl == l && tr == r) return tw;
    int mid = (tl + tr) >> 1;
    if(r <= mid) return query(L, l, r);
    else if(l > mid) return query(R, l, r);
    else {
        return query(L, l, mid) + query(R, 1 + mid, r);
    }
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("data.txt", "r", stdin);
    #endif

    int t, n, a, b, c;
    char s[6];
    scanf("%d", &t);

    rep(Case, 1, t) {
        scanf("%d", &n);
        build(1, 1, n);
        printf("Case %d:\n", Case);
        while(true) {
            scanf("%s", s);
            if(strcmp(s, "End") == 0)
                break;
            if(strcmp(s, "Add") == 0) {
                scanf("%d%d", &a, &b);
                add(1, a, b);
            }
            else if(strcmp(s, "Sub") == 0) {
                scanf("%d%d", &a, &b);
                add(1, a, -b);
            }
            else {
                scanf("%d%d", &a, &b);
                printf("%d\n", query(1, a, b));
            }
        }

    }

    return 0;
}

B - I Hate It

 HDU - 1754

很多学校流行一种比较的习惯。老师们很喜欢询问,从某某到某某当中,分数最高的是多少。 
这让很多学生很反感。 

不管你喜不喜欢,现在需要你做的是,就是按照老师的要求,写一个程序,模拟老师的询问。当然,老师有时候需要更新某位同学的成绩。

Input

本题目包含多组测试,请处理到文件结束。 
在每个测试的第一行,有两个正整数 N 和 M ( 0<N<=200000,0<M<5000 ),分别代表学生的数目和操作的数目。 
学生ID编号分别从1编到N。 
第二行包含N个整数,代表这N个学生的初始成绩,其中第i个数代表ID为i的学生的成绩。 
接下来有M行。每一行有一个字符 C (只取'Q'或'U') ,和两个正整数A,B。 
当C为'Q'的时候,表示这是一条询问操作,它询问ID从A到B(包括A,B)的学生当中,成绩最高的是多少。 
当C为'U'的时候,表示这是一条更新操作,要求把ID为A的学生的成绩更改为B。 

Output

对于每一次询问操作,在一行里面输出最高成绩。

Sample Input

5 6
1 2 3 4 5
Q 1 5
U 3 6
Q 3 4
Q 4 5
U 2 9
Q 1 5

Sample Output

5
6
5
9          

Hint

Huge input,the C function scanf() will work better than cin

单点更新,区间维护最值。

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
#define tl tree[o].l
#define tr tree[o].r
#define ts tree[o].high
#define L o<<1
#define R o<<1|1
#define N 800010
typedef long long ll;

int n, m, a, b, c;
char ch;
struct node{
  int l, r;
  ll high;
}tree[N];

void build(int o, int l, int r)
{
    tl = l; tr = r; ts = 0;
    if(l == r) {
        scanf("%lld", &ts);
        return;
    }
    build(L, l, (l + r)>>1);
    build(R, 1 + ((l + r)>>1), r);
    ts = max(tree[L].high, tree[R].high);
}
void add(int o, int i, ll v)
{
    if(tl == tr) {
        ts = v;
        return;
    }
    ts = max(ts, v);
    int mid = (tl + tr) >> 1;
    if(i <= mid) add(L, i, v);
    else if(i > mid) add(R, i, v);
}
ll query(int o, int l, int r)
{//cout << o <<' '<<l<<' '<<r<<' ' <<tl<<' ' <<tr<<endl;
    if(tl == l && tr == r) return ts;
    int mid = (tl + tr) >> 1;
    if(r <= mid) return query(L, l, r);
    else if(l > mid) return query(R, l, r);
    else {
        return max(query(L, l, mid), query(R, mid + 1, r));
    }
}

int main()
{
    freopen("data.txt", "r", stdin);

    while(~scanf("%d%d", &n, &m)) {
        build(1, 1, n);
        while(m--) {
            cin >> ch;
            if(ch == 'Q'){
                scanf("%d%d", &a, &b);
                printf("%lld\n", query(1, a, b));
            }
            else if(ch == 'U'){
                scanf("%d%d", &a, &b);
                add(1, a, b);
            }

        }
    }

    return 0;
}

C - A Simple Problem with Integers

 POJ - 3468

You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of AaAa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of AaAa+1, ... , Ab.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Output

4
55
9
15

Hint

The sums may exceed the range of 32-bit integers.

区间更新,区间求和,加懒惰标记。

#include <queue>
#include <cstdio>
#include <cstring>
#include <utility>
#include <iostream>
#include <map>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
using namespace std;
typedef pair<int, int> P;
typedef long long ll;
#define N 400010
#define M 3645
const int INF = 0x3f3f3f3f;
const double eps = 1e-5;
const double PI = acos(-1);
#define fi first
#define se second
#define L o<<1
#define R o<<1|1
#define tl tree[o].l
#define tr tree[o].r
#define tw tree[o].w
#define rep(i, lll, nnn) for(int i = (lll); i <= (nnn); i++)

struct Node {
    int l, r, length;
    ll lazy, w;
}tree[N];

void up(int o) {
    tw = tree[L].w + tree[R].w;
}

void down(int o)
{
    if(!tree[o].lazy) return;
    tree[L].w += tree[o].lazy * tree[L].length;
    tree[R].w += tree[o].lazy * tree[R].length;
    tree[L].lazy += tree[o].lazy;
    tree[R].lazy += tree[o].lazy;
    tree[o].lazy = 0;
}

void build(int o, int l, int r)
{
    tl = l; tr = r;
    tree[o].length = r - l + 1;
    tree[o].lazy = 0;
    if(l == r) {
        scanf("%lld", &tw);
        return;
    }
    int mid = (l + r) >> 1;
    build(L, l, mid);
    build(R, 1 + mid, r);
    up(o);
}

ll query(int o, int l, int r)
{
    if(tl == l && tr == r) return tw;
    down(o);
    int mid = (tl + tr) >> 1;
    if(r <= mid) return query(L, l, r);
    else if(l > mid) return query(R, l, r);
    else {
        return query(L, l, mid) + query(R, 1 + mid, r);
    }
}

void add(int o, int l, int r, ll w)
{
    if(tl == l && tr == r) {
        tw += w * tree[o].length;
        tree[o].lazy += w;
        return;
    }
    down(o);
    int mid = (tl + tr) >> 1;
    if(r <= mid) add(L, l, r, w);
    else if(l > mid) add(R, l, r, w);
    else {
        add(L, l, mid, w);
        add(R, 1 + mid, r, w);
    }
    up(o);
}

int n, q, a, b, c;
char s[3];

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("data.txt", "r", stdin);
    #endif

    while(~scanf("%d%d", &n, &q)) {
        build(1, 1, n);

        while(q--) {
            scanf("%s", s);
            if(s[0] == 'Q') {
                scanf("%d%d", &a, &b);
                printf("%lld\n", query(1, a, b));
            }
            else {
                scanf("%d%d%d", &a, &b, &c);
                add(1, a, b, c);
            }
        }
    }

    return 0;
}

D - Mayor's posters

 POJ - 2528

The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral posters at all places at their whim. The city council has finally decided to build an electoral wall for placing the posters and introduce the following rules: 

  • Every candidate can place exactly one poster on the wall. 
  • All posters are of the same height equal to the height of the wall; the width of a poster can be any integer number of bytes (byte is the unit of length in Bytetown). 
  • The wall is divided into segments and the width of each segment is one byte. 
  • Each poster must completely cover a contiguous number of wall segments.


They have built a wall 10000000 bytes long (such that there is enough place for all candidates). When the electoral campaign was restarted, the candidates were placing their posters on the wall and their posters differed widely in width. Moreover, the candidates started placing their posters on wall segments already occupied by other posters. Everyone in Bytetown was curious whose posters will be visible (entirely or in part) on the last day before elections. 
Your task is to find the number of visible posters when all the posters are placed given the information about posters' size, their place and order of placement on the electoral wall. 

Input

The first line of input contains a number c giving the number of cases that follow. The first line of data for a single case contains number 1 <= n <= 10000. The subsequent n lines describe the posters in the order in which they were placed. The i-th line among the n lines contains two integer numbers l i and ri which are the number of the wall segment occupied by the left end and the right end of the i-th poster, respectively. We know that for each 1 <= i <= n, 1 <= l i <= ri <= 10000000. After the i-th poster is placed, it entirely covers all wall segments numbered l i, l i+1 ,... , ri.

Output

For each input data set print the number of visible posters after all the posters are placed. 

The picture below illustrates the case of the sample input. 

Sample Input

1
5
1 4
2 6
8 10
3 4
7 10

Sample Output

4

离散化+逆向思维,这个离散化直接开数组就行了。

#include <queue>
#include <cstdio>
#include <cstring>
#include <utility>
#include <iostream>
#include <map>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
using namespace std;
typedef pair<int, int> P;
typedef long long ll;
#define N 10010
#define M 3645
const int INF = 0x3f3f3f3f;
const double eps = 1e-5;
const double PI = acos(-1);
#define fi first
#define se second
#define L o<<1
#define R o<<1|1
#define tl tree[o].l
#define tr tree[o].r
#define tc tree[o].cover
#define rep(i, lll, nnn) for(int i = (lll); i <= (nnn); i++)

int hash[10000001], num[N * 2];
int t, n, cnt, ans;
struct Post {
  int l, r;
}p[N];

struct Node {
    int l, r;
    bool cover;
}tree[N * 8];

void build(int o, int l, int r)
{
    tl = l; tr = r; tc = false;
    if(l == r) return;
    int mid = (l + r) >> 1;
    build(L, l, mid);
    build(R, 1 + mid, r);
}

bool add(int o, int l, int r)
{
    if(tc) return false;
    if(tl == l && tr == r) return tc = true;

    int mid = (tl + tr) >> 1;
    bool flag = false;
    if(r <= mid) flag = add(L, l, r);
    else if(l > mid) flag = add(R, l, r);
    else {
        bool f1 = add(L, l, mid);
        bool f2 = add(R, 1 + mid, r);
        flag = f1 || f2;
    }
    if(tree[L].cover && tree[R].cover) tc = true;
    return flag;
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("data.txt", "r", stdin);
    #endif

    scanf("%d", &t);

    while(t--) {
        scanf("%d", &n);
        ans = cnt = 0;

        rep(i, 0, n - 1) scanf("%d%d", &p[i].l, &p[i].r);
        rep(i, 0, n - 1) {
            num[cnt++] = p[i].l;
            num[cnt++] = p[i].r;
        }
        sort(num, num + cnt);
        cnt = unique(num, num + cnt) - num;
        rep(i, 0, cnt - 1) {
            hash[num[i]] = i + 1;
        }

        build(1, 1, cnt);

        for(int i = n - 1; i >= 0; i--) {
            if(add(1, hash[p[i].l], hash[p[i].r])) ans++;
        }

        printf("%d\n", ans);
    }

    return 0;
}

E - Just a Hook

 HDU - 1698

In the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of the heroes. The hook is made up of several consecutive metallic sticks which are of the same length. 



Now Pudge wants to do some operations on the hook. 

Let us number the consecutive metallic sticks of the hook from 1 to N. For each operation, Pudge can change the consecutive metallic sticks, numbered from X to Y, into cupreous sticks, silver sticks or golden sticks. 
The total value of the hook is calculated as the sum of values of N metallic sticks. More precisely, the value for each kind of stick is calculated as follows: 

For each cupreous stick, the value is 1. 
For each silver stick, the value is 2. 
For each golden stick, the value is 3. 

Pudge wants to know the total value of the hook after performing the operations. 
You may consider the original hook is made up of cupreous sticks. 

Input

The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 10 cases. 
For each case, the first line contains an integer N, 1<=N<=100,000, which is the number of the sticks of Pudge’s meat hook and the second line contains an integer Q, 0<=Q<=100,000, which is the number of the operations. 
Next Q lines, each line contains three integers X, Y, 1<=X<=Y<=N, Z, 1<=Z<=3, which defines an operation: change the sticks numbered from X to Y into the metal kind Z, where Z=1 represents the cupreous kind, Z=2 represents the silver kind and Z=3 represents the golden kind. 

Output

For each case, print a number in a line representing the total value of the hook after the operations. Use the format in the example. 

Sample Input

1
10
2
1 5 2
5 9 3

Sample Output

Case 1: The total value of the hook is 24.

区间更新+懒惰标记

#include <queue>
#include <cstdio>
#include <cstring>
#include <utility>
#include <iostream>
#include <map>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
using namespace std;
typedef pair<int, int> P;
typedef long long ll;
#define N 400010
#define M 3645
const int INF = 0x3f3f3f3f;
const double eps = 1e-5;
const double PI = acos(-1);
#define fi first
#define se second
#define L o<<1
#define R o<<1|1
#define tl tree[o].l
#define tr tree[o].r
#define tw tree[o].w
#define rep(i, lll, nnn) for(int i = (lll); i <= (nnn); i++)

int t, n, q, a, b, c;

struct Node {
  int l, r, lazy;
  ll w;
}tree[N];

void up(int o) {
    tw = tree[L].w + tree[R].w;
}

void down(int o)
{
    if(!tree[o].lazy) return;
    tree[L].w = (tree[L].r - tree[L].l + 1) * tree[o].lazy;
    tree[R].w = (tree[R].r - tree[R].l + 1) * tree[o].lazy;
    tree[L].lazy = tree[o].lazy;
    tree[R].lazy = tree[o].lazy;
    tree[o].lazy = 0;
}

void build(int o, int l, int r)
{
    tl = l; tr = r; tree[o].lazy = 0;
    if(l == r) {
        tw = 1; return;
    }
    int mid = (l + r) >> 1;
    build(L, l, mid);
    build(R, 1 + mid, r);
    up(o);
}

void change(int o, int l, int r, ll w)
{
    if(tl == l && tr == r) {
        tw = (r - l + 1) * w;
        tree[o].lazy = w;
        return;
    }
    down(o);
    int mid = (tl + tr) >> 1;
    if(r <= mid) change(L, l, r, w);
    else if(l > mid) change(R, l, r, w);
    else {
        change(L, l, mid, w);
        change(R, 1 + mid, r, w);
    }
    up(o);
}

ll query(int o, int l, int r)
{
    if(tl == l && tr == r) return tw;
    down(o);
    int mid = (tl + tr) >> 1;
    if(r <= mid) return query(L, l, r);
    else if(l > mid) return query(R, l, r);
    else {
        return query(L, l, mid) + query(R, 1 + mid, r);
    }
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("data.txt", "r", stdin);
    #endif

    scanf("%d", &t);

    rep(Case, 1, t) {
        scanf("%d%d", &n, &q);
        build(1, 1, n);

        while(q--) {
            scanf("%d%d%d", &a, &b, &c);
            change(1, a, b, c);
        }

        printf("Case %d: The total value of the hook is %lld.\n", Case, query(1, 1, n));
    }

    return 0;
}

F - Count the Colors

 ZOJ - 1610

Painting some colored segments on a line, some previously painted segments may be covered by some the subsequent ones.

Your task is counting the segments of different colors you can see at last.

 

Input



The first line of each data set contains exactly one integer n, 1 <= n <= 8000, equal to the number of colored segments.

 

Each of the following n lines consists of exactly 3 nonnegative integers separated by single spaces:

x1 x2 c

x1 and x2 indicate the left endpoint and right endpoint of the segment, c indicates the color of the segment.

All the numbers are in the range [0, 8000], and they are all integers.

Input may contain several data set, process to the end of file.

 

Output



Each line of the output should contain a color index that can be seen from the top, following the count of the segments of this color, they should be printed according to the color index.

 

If some color can't be seen, you shouldn't print it.

Print a blank line after every dataset.

 

Sample Input



5
0 4 4
0 3 1
3 4 2
0 2 2
0 2 3
4
0 1 1
3 4 1
1 3 2
1 3 1
6
0 1 0
1 2 1
2 3 1
1 2 0
2 3 0
1 2 1

 

 

Sample Output



1 1
2 1
3 1

 

1 1

0 2
1 1

看了看代码发现我是把大区间的值赋值为-1,现在写肯定是加ok标记和up操作,count的时候就不用pre来搞了。

#include <queue>
#include <cstdio>
#include <cstring>
#include <utility>
#include <iostream>
#include <map>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
using namespace std;
typedef pair<int, int> P;
typedef long long ll;
#define N 8010
#define M 3645
const int INF = 0x3f3f3f3f;
const double eps = 1e-5;
const double PI = acos(-1);
#define fi first
#define se second
#define L o<<1
#define R o<<1|1
#define tl tree[o].l
#define tr tree[o].r
#define tw tree[o].w
#define rep(i, lll, nnn) for(int i = (lll); i <= (nnn); i++)

int n, a, b, c;
int cnt[N];

struct Node {
    int l, r, w;
}tree[N * 4];

void build(int o, int l, int r)
{
    tl = l; tr = r; tw = 0;
    if(l == r) return;
    int mid = (l + r) >> 1;
    build(L, l, mid);
    build(R, 1 + mid, r);
}

void change(int o, int l, int r, int w)
{
  //  if(w == 2) cout <<l <<' ' <<r << ' ' <<tl<<' ' <<tr<<endl;
    if(tw == w) return;
    if(l <= tl && tr <= r) {
        tw = w;
        return;
    }
    if(tw > 0) {
        tree[L].w = tree[R].w = tw;
        tw = -1;
    }

    int mid = (tl + tr) >> 1;
    if(r <= mid) change(L, l, r, w);
    else if(l > mid) change(R, l, r, w);
    else {
        change(L, l, mid, w);
        change(R, 1 + mid, r, w);
    }
    tw = -1;
}

int pre;
void count(int o)
{
    //if(tr < 100)cout << tl <<' ' <<tr <<' '<<tw <<endl;
    if(tw == 0) {
        pre = 0;
        return;
    }

    if(tw > 0) {
        if(tw != pre) {
            pre = tw;
            cnt[tw]++;
        }
        return;
    }

    if(tl != tr) {
        count(L);
        count(R);
    }
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("data.txt", "r", stdin);
    #endif

    while(~scanf("%d", &n)) {
        memset(cnt, 0, sizeof cnt);
        build(1, 1, 8001);

        rep(i, 1, n) {
            scanf("%d%d%d", &a, &b, &c);
            change(1, a + 1, b, c + 1);
        }
        pre = 0;
        count(1);

        rep(i, 1, 8001) if(cnt[i]) {
            printf("%d %d\n", i - 1, cnt[i]);
        }
        putchar('\n');
    }

    return 0;
}

G - Balanced Lineup

 POJ - 3264

For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.

Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.

Input

Line 1: Two space-separated integers, N and Q
Lines 2.. N+1: Line i+1 contains a single integer that is the height of cow i 
Lines N+2.. NQ+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.

Output

Lines 1.. Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.

Sample Input

6 3
1
7
3
4
2
5
1 5
4 6
2 2

Sample Output

6
3
0

差值需要维护区间最值。

#include <queue>
#include <cstdio>
#include <cstring>
#include <utility>
#include <iostream>
#include <map>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
using namespace std;
typedef pair<int, int> P;
typedef long long ll;
#define N 200010
#define M 3645
const int INF = 0x3f3f3f3f;
const double eps = 1e-5;
const double PI = acos(-1);
#define fi first
#define se second
#define L o<<1
#define R o<<1|1
#define tl tree[o].l
#define tr tree[o].r
#define ta tree[o].a
#define tb tree[o].b
#define rep(i, lll, nnn) for(int i = (lll); i <= (nnn); i++)

struct Node {
  int l, r;
  int a, b;
}tree[N];
int n, q, maxh, minh;

void build(int o, int l, int r)
{
    tl = l; tr = r;
    if(l == r) {
        scanf("%d", &ta);
        tb = ta;
        return;
    }
    int mid = (l + r) >> 1;
    build(L, l, mid);
    build(R, 1 + mid, r);
    ta = min(tree[L].a, tree[R].a);
    tb = max(tree[L].b, tree[R].b);
}

void query(int o, int l, int r)
{
    if(tl == l && tr == r) {
        minh = min(minh, ta);
        maxh = max(maxh, tb);
        return;
    }

    int mid = (tl + tr) >> 1;
    if(r <= mid) query(L, l, r);
    else if(l > mid) query(R, l, r);
    else {
        query(L, l, mid);
        query(R, 1 + mid, r);
    }
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("data.txt", "r", stdin);
    #endif

    while(~scanf("%d%d", &n, &q)) {
        build(1, 1, n);

        while(q--) {
            int a, b;
            scanf("%d%d", &a, &b);
            maxh = 0; minh = INF;
            query(1, a, b);
            printf("%d\n", maxh - minh);
        }
    }

    return 0;
}

H - Can you answer these queries?

 HDU - 4027

A lot of battleships of evil are arranged in a line before the battle. Our commander decides to use our secret weapon to eliminate the battleships. Each of the battleships can be marked a value of endurance. For every attack of our secret weapon, it could decrease the endurance of a consecutive part of battleships by make their endurance to the square root of it original value of endurance. During the series of attack of our secret weapon, the commander wants to evaluate the effect of the weapon, so he asks you for help. 
You are asked to answer the queries that the sum of the endurance of a consecutive part of the battleship line. 

Notice that the square root operation should be rounded down to integer.

Input

The input contains several test cases, terminated by EOF. 
  For each test case, the first line contains a single integer N, denoting there are N battleships of evil in a line. (1 <= N <= 100000) 
  The second line contains N integers Ei, indicating the endurance value of each battleship from the beginning of the line to the end. You can assume that the sum of all endurance value is less than 2 63. 
  The next line contains an integer M, denoting the number of actions and queries. (1 <= M <= 100000) 
  For the following M lines, each line contains three integers T, X and Y. The T=0 denoting the action of the secret weapon, which will decrease the endurance value of the battleships between the X-th and Y-th battleship, inclusive. The T=1 denoting the query of the commander which ask for the sum of the endurance value of the battleship between X-th and Y-th, inclusive. 

Output

For each test case, print the case number at the first line. Then print one line for each query. And remember follow a blank line after each test case.

Sample Input

10
1 2 3 4 5 6 7 8 9 10
5
0 1 10
1 1 10
1 1 5
0 5 8
1 4 8

Sample Output

Case #1:
19
7
6

区间更新而且区间只能单点更新,但是每个点更新的次数是有限的,最多七八次,所以加个标记记录下这个区间是否能更新就行,

#include <queue>
#include <cstdio>
#include <cstring>
#include <utility>
#include <iostream>
#include <map>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
using namespace std;
typedef pair<int, int> P;
typedef long long ll;
#define N 400010
#define M 3645
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1);
#define fi first
#define se second
#define L o<<1
#define R o<<1|1
#define tl tree[o].l
#define tr tree[o].r
#define tw tree[o].w
#define to tree[o].ok
#define rep(i, lll, nnn) for(int i = (lll); i <= (nnn); i++)

int n, m, type, a, b;
struct Node {
  int l, r;
  ll w;
  bool ok;
}tree[N];

void up(int o){
    tw = tree[L].w + tree[R].w;
    to = tree[L].ok || tree[R].ok;
}

void build(int o, int l, int r)
{
    tl = l; tr = r;
    if(l == r) {
        scanf("%lld", &tw);
        if(tw == 1) to = false;
        else to = true;
        return;
    }
    int mid = (l + r) >> 1;
    build(L, l, mid);
    build(R, 1 + mid, r);
    up(o);
}

void change(int o, int l, int r)
{
    if(!to) return;
    if(tl == tr) {
        tw = sqrt((double)tw);
        if(tw <= 1) to = false;
        return;
    }

//    int mid = (tl + tr) >> 1;
//    if(r <= mid) {
//        if(tree[L].ok) change(L, l, r);
//    }
//    else if(l > mid) {
//        if(tree[R].ok) change(R, l, r);
//    }
//    else {
//        if(tree[L].ok) change(L, l, mid);
//        if(tree[R].ok) change(R, 1 + mid, r);
//    }
    int mid = (tl + tr) >> 1;
    if(l <= mid && tree[L].ok) change(L, l, r);
    if(r > mid && tree[R].ok) change(R, l, r);

    up(o);
}

ll query(int o, int l, int r)
{
    if(tl == l && tr == r) return tw;
    int mid = (tl + tr) >> 1;
    if(r <= mid) return query(L, l, r);
    else if(l > mid) return query(R, l, r);
    else {
        return query(L, l, mid) + query(R, 1 + mid, r);
    }
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("data.txt", "r", stdin);
    #endif

    int Case = 1;
    while(~scanf("%d", &n)) {
        build(1, 1, n);

        scanf("%d", &m);
        printf("Case #%d:\n", Case++);
        while(m--) {
            scanf("%d%d%d", &type, &a, &b);
            if(a > b) swap(a, b);
            if(type) {
                printf("%lld\n", query(1, a, b));
            }
            else {
                change(1, a, b);
            }
      //  printf("   %lld\n", query(1, a, b));
        }
        puts("");
    }

    return 0;
}

I - Tunnel Warfare

 HDU - 1540

During the War of Resistance Against Japan, tunnel warfare was carried out extensively in the vast areas of north China Plain. Generally speaking, villages connected by tunnels lay in a line. Except the two at the ends, every village was directly connected with two neighboring ones. 

Frequently the invaders launched attack on some of the villages and destroyed the parts of tunnels in them. The Eighth Route Army commanders requested the latest connection state of the tunnels and villages. If some villages are severely isolated, restoration of connection must be done immediately! 

Input

The first line of the input contains two positive integers n and m (n, m ≤ 50,000) indicating the number of villages and events. Each of the next m lines describes an event. 

There are three different events described in different format shown below: 

D x: The x-th village was destroyed. 

Q x: The Army commands requested the number of villages that x-th village was directly or indirectly connected with including itself. 

R: The village destroyed last was rebuilt. 

Output

Output the answer to each of the Army commanders’ request in order on a separate line. 

Sample Input

7 9
D 3
D 6
D 5
Q 4
Q 5
R
Q 4
R
Q 4

Sample Output

1
0
2
4

第一眼这不是个set吗,然后set真过了。

区间合并的题,第一次写区间合并。

#include <queue>
#include <cstdio>
#include <cstring>
#include <utility>
#include <iostream>
#include <map>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <set>
using namespace std;
typedef pair<int, int> P;
typedef long long ll;
#define N 500100
#define M 3645
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1);
#define fi first
#define se second
#define L o<<1
#define R o<<1|1
#define tl tree[o].l
#define tr tree[o].r
#define tlw tree[o].lw
#define trw tree[o].rw
#define rep(i, lll, nnn) for(int i = (lll); i <= (nnn); i++)

int n, m;
struct Node {
    int l, r;
    int lw, rw;
}tree[N * 4];
bool vis[N];
int a[N], l;

void up(int o)
{
    tlw = tree[L].lw;
    trw = tree[R].rw;
    int mid = (tl + tr) >> 1;
    if(tree[L].lw == tree[L].r - tree[L].l + 1) {
        tlw += tree[R].lw;
    }
    if(tree[R].rw == tree[R].r - tree[R].l + 1) {
        trw += tree[L].rw;
    }
}

void build(int o, int l, int r)
{
    tl = l; tr = r;
    if(l == r) {
        tlw = trw = 1;
        return;
    }
    int mid = (l + r) >> 1;
    build(L, l, mid);
    build(R, 1 + mid, r);
    up(o);
}

void change(int o, int i, bool v)
{
    if(tl == tr) {
        if(v == true) tlw = trw = 1;
        else tlw = trw = 0;
        return;
    }

    int mid = (tl + tr) >> 1;
    if(i <= mid) change(L, i, v);
    else change(R, i, v);

    up(o);
}

int query(int o, int i)
{
//cout << tl << ' ' << tr << ' ' << i << endl;
    if(tl == tr || tlw == tr - tl + 1) return tlw;
    int mid = (tl + tr) >> 1;
    if(i <= mid) {
        if(i >= mid - tree[L].rw + 1) return query(L, i) + query(R, mid + 1);
        return query(L, i);
    }
    if(i <= mid +  tree[R].lw)return query(L, mid) + query(R, i);
    return query(R, i);
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("data.txt", "r", stdin);
    #endif

    int x;
    char s[3];
    while(~scanf("%d%d", &n, &m)) {
        memset(vis, false, sizeof vis);
        l = 0;
        build(1, 1, n);
        while(m--) {
            scanf("%s", s);
            if(s[0] == 'D') {
                scanf("%d", &x);
                vis[x] = true;
                a[l++] = x;
                change(1, x, false);
            }
            else if(s[0] == 'Q') {
                scanf("%d", &x);
                if(vis[x]) {
                    puts("0");
                    continue;
                }
                printf("%d\n", query(1, x));
            }
            else {
                if(l == 0) continue;
                change(1, a[l - 1], true);
                vis[a[l - 1]] = false;
                l--;
            }
        }
    }

    return 0;
}


//set<int> st;
//int a[N], l;
//int n, m, x;
//char s[5];
//
//int main()
//{
//    #ifndef ONLINE_JUDGE
//    freopen("data.txt", "r", stdin);
//    #endif
//
//    set<int>::iterator it1, it2;
//    while(~scanf("%d%d", &n, &m)) {
//        st.clear();
//        st.insert(0);
//        st.insert(n + 1);
//        l = 0;
//        while(m--) {
//            scanf("%s", &s);
//            if(s[0] == 'D') {
//                scanf("%d", &x);
//                a[l++] = x;
//                st.insert(x);
//            }
//            else if(s[0] == 'Q') {
//                scanf("%d", &x);
//                if(st.find(x) != st.end()) {
//                    puts("0"); continue;
//                }
//                it1 = it2 = st.lower_bound(x);
//                int r, l;
//                r = *it2;
//                l = *(--it1);
//                printf("%d\n", r - l - 1);
//            }
//            else {
//                if(l == 0) continue;
//                st.erase(a[--l]);
//            }
//        }
//    }
//
//    return 0;
//}

J - Assign the task

 HDU - 3974

There is a company that has N employees(numbered from 1 to N),every employee in the company has a immediate boss (except for the leader of whole company).If you are the immediate boss of someone,that person is your subordinate, and all his subordinates are your subordinates as well. If you are nobody's boss, then you have no subordinates,the employee who has no immediate boss is the leader of whole company.So it means the N employees form a tree. 

The company usually assigns some tasks to some employees to finish.When a task is assigned to someone,He/She will assigned it to all his/her subordinates.In other words,the person and all his/her subordinates received a task in the same time. Furthermore,whenever a employee received a task,he/she will stop the current task(if he/she has) and start the new one. 

Write a program that will help in figuring out some employee’s current task after the company assign some tasks to some employee.

Input

The first line contains a single positive integer T( T <= 10 ), indicates the number of test cases. 

For each test case: 

The first line contains an integer N (N ≤ 50,000) , which is the number of the employees. 

The following N - 1 lines each contain two integers u and v, which means the employee v is the immediate boss of employee u(1<=u,v<=N). 

The next line contains an integer M (M ≤ 50,000). 

The following M lines each contain a message which is either 

"C x" which means an inquiry for the current task of employee x 

or 

"T x y"which means the company assign task y to employee x. 

(1<=x<=N,0<=y<=10^9)

Output

For each test case, print the test case number (beginning with 1) in the first line and then for every inquiry, output the correspond answer per line.

Sample Input

1 
5 
4 3 
3 2 
1 3 
5 2 
5 
C 3 
T 2 1
 C 3 
T 3 2 
C 3

Sample Output

Case #1:
-1 
1 
2

emmmmmm百度下来叫线段树维护dfs序,,先dfs出每个节点的子孙有多少个以及该节点在dfs树上和线段树的位置,再对线段操作。

#include <queue>
#include <cstdio>
#include <cstring>
#include <utility>
#include <iostream>
#include <map>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <set>
using namespace std;
typedef pair<int, int> P;
typedef long long ll;
#define N 50010
#define M 3645
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1);
#define fi first
#define se second
#define L o<<1
#define R o<<1|1
#define tl tree[o].l
#define tr tree[o].r
#define tw tree[o].w
#define tz tree[o].lazy
#define rep(i, lll, nnn) for(int i = (lll); i <= (nnn); i++)

vector<int> G[N];
int root, deg[N];
int pos[N], cnt[N], cntu;

int dfs(int fa)
{
    int res = 1;
    pos[fa] = cntu++;
    for(int i = 0; i < G[fa].size(); i++) {
        res += dfs(G[fa][i]);
    }
    return cnt[fa] = res;
}

int n, m, x, y;
char s[3];

struct Node {
  int l, r, lazy, w;
}tree[N * 4];

void down(int o)
{
    if(tz == -1) return;
    tree[L].w = tree[L].lazy = tz;
    tree[R].w = tree[R].lazy = tz;
    tz = -1;
}

void build(int o, int l, int r)
{
    tl = l; tr = r; tz = tw = -1;
    if(l == r) return;

    int mid = (l + r) >> 1;
    build(L, l, mid);
    build(R, 1 + mid, r);
}

void change(int o, int l, int r, int v)
{
    if(tl == tr) {
        tw = tz = v;
        return;
    }
    down(o);
    int mid = (tl + tr) >> 1;
    if(r <= mid) change(L, l, r, v);
    else if(l > mid) change(R, l, r, v);
    else {
        change(L, l, mid, v);
        change(R, 1 + mid, r, v);
    }
}

int query(int o, int i)
{
    if(tl == tr) return tw;
    down(o);
    int mid = (tl + tr) >> 1;
    if(i <= mid) return query(L, i);
    else return query(R, i);
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("data.txt", "r", stdin);
    #endif

    int t;
    scanf("%d", &t);

    rep(Case, 1, t) {
        scanf("%d", &n);
        rep(i, 1, n) G[i].clear();
        memset(deg, 0, sizeof deg);

        rep(i, 1, n - 1) {
            int u, v;
            scanf("%d%d", &u, &v);
            G[v].push_back(u);
            deg[u]++;
        }
        rep(i, 1, n) if(deg[i] == 0) {
            root = i; break;
        }

        cntu = 1;
        dfs(root);

        build(1, 1, n);

        printf("Case #%d:\n", Case);
        scanf("%d", &m);
        while(m--) {
            scanf("%s", s);
            if(s[0] == 'C') {
                scanf("%d", &x);
                printf("%d\n", query(1, pos[x]));
            }
            else {
                scanf("%d%d", &x, &y);
                change(1, pos[x], pos[x] + cnt[x] - 1, y);
            }
        }
    }

    return 0;
}

K - Transformation

 HDU - 4578

Yuanfang is puzzled with the question below: 
There are n integers, a 1, a 2, …, a n. The initial values of them are 0. There are four kinds of operations. 
Operation 1: Add c to each number between a x and a y inclusive. In other words, do transformation a k<---a k+c, k = x,x+1,…,y. 
Operation 2: Multiply c to each number between a x and a y inclusive. In other words, do transformation a k<---a k×c, k = x,x+1,…,y. 
Operation 3: Change the numbers between a x and a y to c, inclusive. In other words, do transformation a k<---c, k = x,x+1,…,y. 
Operation 4: Get the sum of p power among the numbers between a x and a y inclusive. In other words, get the result of a x p+a x+1 p+…+a y p. 
Yuanfang has no idea of how to do it. So he wants to ask you to help him. 

Input

There are no more than 10 test cases. 
For each case, the first line contains two numbers n and m, meaning that there are n integers and m operations. 1 <= n, m <= 100,000. 
Each the following m lines contains an operation. Operation 1 to 3 is in this format: "1 x y c" or "2 x y c" or "3 x y c". Operation 4 is in this format: "4 x y p". (1 <= x <= y <= n, 1 <= c <= 10,000, 1 <= p <= 3) 
The input ends with 0 0. 

Output

For each operation 4, output a single integer in one line representing the result. The answer may be quite large. You just need to calculate the remainder of the answer when divided by 10007.

Sample Input

5 5
3 3 5 7
1 2 4 4
4 1 5 2
2 2 5 8
4 3 5 3
0 0

Sample Output

307
7489

三种区间更新,区间只要维护该区间每个点的值就行了。

#include <queue>
#include <cstdio>
#include <cstring>
#include <utility>
#include <iostream>
#include <map>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
using namespace std;
typedef pair<int, int> P;
typedef long long ll;
#define N 100010
#define M 3645
const int INF = 0x3f3f3f3f;
const double eps = 1e-5;
const double PI = acos(-1);
const int mod = 10007;
#define fi first
#define se second
#define L o<<1
#define R o<<1|1
#define tl tree[o].l
#define tr tree[o].r
#define tw tree[o].w
#define to tree[o].ok
#define rep(i, lll, nnn) for(int i = (lll); i <= (nnn); i++)

int n, m;
struct Node {
  int l, r, w;
  bool ok;
}tree[N * 4];

int power(int a, int b)
{
    int res = 1;
    while(b) {
        if(b & 1) res = (res * a) % mod;
        a = (a * a) % mod;
        b >>= 1;
    }
    return res;
}

void build(int o, int l, int r)
{
    tl = l; tr = r; to = true; tw  = 0;
    if(l == r) return;
    int mid = (l + r) >> 1;
    build(L, l, mid);
    build(R, 1 + mid, r);
}

void up(int o)
{
    if(!tree[L].ok || !tree[R].ok) {
        to = false; return;
    }
    if(tree[L].w != tree[R].w) {
        to = false; return;
    }
    tw = tree[L].w;
    to = true;
}

void down(int o)
{
    if(!to) return;
    tree[L].w = tree[R].w = tw;
    tree[L].ok = tree[R].ok = true;
}

void change(int o, int p, int l, int r, int v)
{
    if(tl == l && tr == r && to) {
        if(p == 1) tw = (tw + v) % mod;
        else if(p == 2) tw = (tw * v) % mod;
        else tw = v % mod;
        return;
    }
    down(o);
    int mid = (tl + tr) >> 1;
    if(r <= mid) change(L, p, l, r, v);
    else if(l > mid) change(R, p, l, r, v);
    else {
        change(L, p, l, mid, v);
        change(R, p, 1 + mid, r, v);
    }
    up(o);
}

int query(int o, int l, int r, int p)
{
    if(tl == l && tr == r && to) {
        int res = power(tw, p);
        res = (res * (r - l + 1)) % mod;
        return res;
    }
    down(o);
    int mid = (tl + tr) >> 1;
    if(r <= mid) return query(L, l, r, p);
    else if(l > mid) return query(R, l, r, p);
    else {
        return (query(L, l, mid, p) + query(R, 1 + mid, r, p)) % mod;
    }
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("data.txt", "r", stdin);
    #endif

    int a, b, c, d;
    while(~scanf("%d%d", &n, &m) && n + m) {
        build(1, 1, n);

        while(m--) {
            scanf("%d%d%d%d", &a, &b, &c, &d);
            if(a == 4) {
                printf("%d\n", query(1, b, c, d));
            }
            else {
                change(1, a, b, c, d);
            }
        }
    }

    return 0;
}

L - Vases and Flowers

 HDU - 4614

Alice is so popular that she can receive many flowers everyday. She has N vases numbered from 0 to N-1. When she receive some flowers, she will try to put them in the vases, one flower in one vase. She randomly choose the vase A and try to put a flower in the vase. If the there is no flower in the vase, she will put a flower in it, otherwise she skip this vase. And then she will try put in the vase A+1, A+2, ..., N-1, until there is no flower left or she has tried the vase N-1. The left flowers will be discarded. Of course, sometimes she will clean the vases. Because there are too many vases, she randomly choose to clean the vases numbered from A to B(A <= B). The flowers in the cleaned vases will be discarded.

Input

  The first line contains an integer T, indicating the number of test cases. 
  For each test case, the first line contains two integers N(1 < N < 50001) and M(1 < M < 50001). N is the number of vases, and M is the operations of Alice. Each of the next M lines contains three integers. The first integer of one line is K(1 or 2). If K is 1, then two integers A and F follow. It means Alice receive F flowers and try to put a flower in the vase A first. If K is 2, then two integers A and B follow. It means the owner would like to clean the vases numbered from A to B(A <= B).

Output

  For each operation of which K is 1, output the position of the vase in which Alice put the first flower and last one, separated by a blank. If she can not put any one, then output 'Can not put any one.'. For each operation of which K is 2, output the number of discarded flowers. 
   Output one blank line after each test case.

Sample Input

2
10 5
1 3 5
2 4 5
1 1 8
2 3 6
1 8 8
10 6
1 2 5
2 3 4
1 0 8
2 2 5
1 4 4
1 2 3

Sample Output

[pre]3 7
2
1 9
4
Can not put any one.

2 6
2
0 9
4
4 5
2 3

[/pre]

区间单点更新,ok标记该区间是否一样,w记录区间里花的数量。

#include <queue>
#include <cstdio>
#include <cstring>
#include <utility>
#include <iostream>
#include <map>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
using namespace std;
typedef pair<int, int> P;
typedef long long ll;
#define N 50010
#define M 3645
const int INF = 0x3f3f3f3f;
const double eps = 1e-5;
const double PI = acos(-1);
const int mod = 10007;
#define fi first
#define se second
#define L o<<1
#define R o<<1|1
#define tl tree[o].l
#define tr tree[o].r
#define tw tree[o].w
#define to tree[o].ok
#define rep(i, lll, nnn) for(int i = (lll); i <= (nnn); i++)

int remain, a, b, cnt;
struct Node {
  int l, r, w;
  bool ok;
}tree[N * 4];

void build(int o, int l, int r)
{
    tl = l; tr = r; tw = 0; to = true;
    if(l == r) return;
    int mid = (tl + tr) >> 1;
    build(L, l, mid);
    build(R, 1 + mid, r);
}

void up(int o)
{
    if(!tree[L].ok || !tree[R].ok) {
        to = false; return;
    }
    if(tree[L].w > 0 && tree[R].w == 0) {
        to = false; return;
    }
    if(tree[R].w > 0 && tree[L].w == 0) {
        to = false; return;
    }
    to = true;
    tw = tree[L].w + tree[R].w;
}

void down(int o)
{
    if(to) {
        if(tw == 0) tree[L].w = tree[R].w = 0;
        else {
            tree[L].w = tree[L].r - tree[L].l + 1;
            tree[R].w = tree[R].r - tree[R].l + 1;
        }
        tree[L].ok = tree[R].ok = to;
    }
}

void change(int o, int l, int r, bool v)
{
    if(remain <= 0) return;
//cout << tl << ' ' << tr << ' ' << l << ' ' <<r << ' ' << to << endl;
    if(tl == l && tr == r && to) {
        if(v) {
            if(tw == 0) {
                if(remain >= tr - tl + 1) {
                    tw = tr - tl + 1;
                    remain -= tr - tl + 1;
                    if(a == -1) a = tl;
                    b = max(b, tr);
                    return;
                }
            }
            else {
                return;
            }
        }
        else {
            cnt += tw;
            tw = 0;
            to = true;
            return;
        }
    }
    down(o);
    int mid = (tl + tr) >> 1;
    if(r <= mid) change(L, l, r, v);
    else if(l > mid) change(R, l, r, v);
    else {
        change(L, l, mid, v);
        change(R, 1 + mid, r, v);
    }
    up(o);
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("data.txt", "r", stdin);
    #endif

    int t, k, x, y, n, m;
    scanf("%d", &t);
    while(t--) {
        scanf("%d%d", &n, &m);
        build(1, 1, n);
    //cout << "a\n";
        while(m--) {
            scanf("%d%d%d", &k, &x, &y);

            if(k == 1) {
                remain = y;
                a = b = -1;
                change(1, x + 1, n, true);
                if(a == -1) puts("Can not put any one.");
                else printf("%d %d\n", a - 1, b - 1);
            }
            else {
                cnt = 0;
                remain = 1;
                change(1, x + 1, y + 1, false);
                printf("%d\n", cnt);
            }
        }
        putchar('\n');
    }

    return 0;
}

 

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