线段树专题

线段树的风格是跟NotOnlySuccess学的

单点更新:最最基础的线段树,只更新叶子节点,然后把信息用PushUP(int r)这个函数更新上来

1、hdu1166 敌兵布阵

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

分析：

1.Add i,j sum[i]+=j

2.Sub i,j sum[i]-=j

3.Query i,j sum[i]+sum[i+1]....sum[j]

最简单的单点更新+区间求和用来入门还是很不错的。

#include <iostream>
#include <cstdio>
#include <cstring>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define maxlen 50010
using namespace std;
int sum[maxlen<<2];
void pushup(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void build(int l,int r,int rt)
{
    if(l==r)
    {
        scanf("%d",&sum[rt]);
        return ;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    pushup(rt);
}
void update(int p,int add,int l,int r,int rt)
{
    if(l==r)
    {
        sum[rt]+=add;
        return;
    }
    int m=(l+r)>>1;
    if(p<=m)
        update(p,add,lson);
    else
        update(p,add,rson);
    pushup(rt);
}
int query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)//查询区间大于当前区间直接返回根节点值
        return sum[rt];
    int m=(l+r)>>1;
    int ans=0;
    if(L<=m)
        ans+=query(L,R,lson);
    if(R>m)
        ans+=query(L,R,rson);
    return ans;
}
int main ()
{
    int t,n;
    char cmd[10];
    scanf("%d",&t);
    for(int Case=1;Case<=t;++Case)
    {
        printf("Case %d:
",Case);
        scanf("%d",&n);
        build(1,n,1);
        while(scanf("%s",&cmd))
        {
            if(cmd[0]=='E')
                break;
            int x,y;
            scanf("%d%d",&x,&y);
            if(cmd[0]=='Q')
                printf("%d
",query(x,y,1,n,1));
            else if(cmd[0]=='S')
                update(x,-y,1,n,1);
            else
                update(x,y,1,n,1);
        }
    }
}

2、hdu1754 I Hate It

Problem Description

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

分析:

1.U i,j sum[i]=j

2.Q i,j max(i,i+,1...,j)

最简单的单点更新+区间求最大值，跟基本上面一样，只是把维护的变为区间最大值。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define maxlen 200010
using namespace std;
int sum[maxlen<<2];
void pushup(int rt)
{
    sum[rt]=max(sum[rt<<1],sum[rt<<1|1]);
}
void build(int l,int r,int rt)
{
    if(l==r)
    {
        scanf("%d",&sum[rt]);
        return ;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    pushup(rt);
}
void update(int p,int add,int l,int r,int rt)
{
    if(l==r)
    {
        sum[rt]=add;
        return;
    }
    int m=(l+r)>>1;
    if(p<=m)
        update(p,add,lson);
    else
        update(p,add,rson);
    pushup(rt);
}
int query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)
        return sum[rt];
    int m=(l+r)>>1;
    int ans=0;
    if(L<=m)
        ans=max(ans,query(L,R,lson));
    if(R>m)
        ans=max(ans,query(L,R,rson));
    return ans;
}
int main ()
{
    int t,n,m;
    char cmd[2];
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        build(1,n,1);
        while(m--)
        {
            int x,y;
            scanf("%s%d%d",cmd,&x,&y);
            if(cmd[0]=='Q')
                printf("%d
",query(x,y,1,n,1));
            else
                update(x,y,1,n,1);
        }
    }
}

3、hdu1394 Minimum Inversion Number

Problem Description

The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.

For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following:

a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
a2, a3, ..., an, a1 (where m = 1)
a3, a4, ..., an, a1, a2 (where m = 2)
...
an, a1, a2, ..., an-1 (where m = n-1)

You are asked to write a program to find the minimum inversion number out of the above sequences.

 

Input

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.

 

Output

For each case, output the minimum inversion number on a single line.

 

Sample Input

10

1 3 6 9 0 8 5 7 4 2

 

Sample Output

16

题目大意：给你n个数（0~n-1）组成一个序列，每次把第一个数移到最后去，这样可以得到n个序列，问这些序列中最小的逆序数。

分析：首先来说一下逆序数的求法

例如要求x的逆序数只需要访问（x+1，n）段有多少个数，就是x的逆序数。还有就是求最小逆序数的时候有个巧妙的想法，当把x放入数组的后面，此时的逆序数应该为x没放入最后面之前的逆序总数加上（n-x）再减去（x-1）；sum = sum+(n-x[i])-(x[i]-1)。

先上一个暴力的，复杂度O(n^2) 300ms+

#include <iostream>
#include <cstdio>
#define maxlen 5010
using namespace std;
int s[maxlen];
int n,ans,cnt;
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        cnt=0,ans=99999999;
        for(int i=1;i<=n;i++)
            scanf("%d",&s[i]);
        for(int i=1;i<=n;i++)
            for(int j=i+1;j<=n;j++)
               if(s[i]>s[j]) cnt++;
        for(int i=1;i<=n;i++)
        {
            cnt=cnt-s[i]+n-s[i]-1;
            ans=min(cnt,ans);
        }
        printf("%d
",ans);
    }
    return 0;
}

线段树 复杂度O(nlogn) 30ms+

单点更新+区间求和

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define maxlen 5010
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
int sum[maxlen<<2];
int s[maxlen];
void pushup(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void build(int l,int r,int rt)
{
    sum[rt]=0;
    if(l==r)
        return ;
    int m=(l+r)>>1;
    build(lson);
    build(rson);
}
void update(int p,int l,int r,int rt)
{
    if(l==r)
    {
        sum[rt]++;
        return ;
    }
    int m=(l+r)>>1;
    if(p<=m)
        update(p,lson);
    else
        update(p,rson);
    pushup(rt);
}
int query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)
        return sum[rt];
    int m=(l+r)>>1;
    int ans=0;
    if(L<=m)
        ans+=query(L,R,lson);
    if (R>m)
        ans+=query(L,R,rson);
    return ans;
}
using namespace std;
int main ()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        build(0,n-1,1);
        int sum=0;
        for(int i=0; i<n; ++i)
        {
            scanf("%d",&s[i]);
            sum+=query(s[i],n-1,0,n-1,1);
            update(s[i],0,n-1,1);
        }
        int ans=sum;
        for(int i=0; i<n; ++i)
        {
            sum+=n-s[i]-s[i]-1;
            ans=min(ans,sum);
        }
        printf("%d
",ans);
    }
}

4、hdu2795 Billboard

Problem Description

At the entrance to the university, there is a huge rectangular billboard of size h*w (h is its height and w is its width). The board is the place where all possible announcements are posted: nearest programming competitions, changes in the dining room menu, and other important information.

On September 1, the billboard was empty. One by one, the announcements started being put on the billboard.

Each announcement is a stripe of paper of unit height. More specifically, the i-th announcement is a rectangle of size 1 * wi.

When someone puts a new announcement on the billboard, she would always choose the topmost possible position for the announcement. Among all possible topmost positions she would always choose the leftmost one.

If there is no valid location for a new announcement, it is not put on the billboard (that's why some programming contests have no participants from this university).

Given the sizes of the billboard and the announcements, your task is to find the numbers of rows in which the announcements are placed.

 

Input

There are multiple cases (no more than 40 cases).

The first line of the input file contains three integer numbers, h, w, and n (1 <= h,w <= 10^9; 1 <= n <= 200,000) - the dimensions of the billboard and the number of announcements.

Each of the next n lines contains an integer number wi (1 <= wi <= 10^9) - the width of i-th announcement.

 

Output

For each announcement (in the order they are given in the input file) output one number - the number of the row in which this announcement is placed. Rows are numbered from 1 to h, starting with the top row. If an announcement can't be put on the billboard, output "-1" for this announcement.

 

Sample Input

3 5 5

2

4

3

3

3

 

Sample Output

1

2

1

3

-1

题目大意：给你h*w的矩形，现在在上面贴1*L的告示，每次优先选择最左上的位置，输出输在的行号，如果不能贴输出-1

分析：

利用线段树可以求区间的最大值;
将位置即h用来建树(h<=n,大了没有意义);
树中存储的为该位置还拥有的空间; （没贴一次减去L）
若左子树的最大值大于他,就查询左子树，否则查询右子树;

线段树 ，单点更新+区间最大值 2500ms+

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define maxlen 200010
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
int sum[maxlen<<2];
int h,w,n;
void pushup(int rt)
{
    sum[rt]=max(sum[rt<<1],sum[rt<<1|1]);
}
void build(int l,int r,int rt)
{
    sum[rt]=w;
    if(l==r)
        return ;
    int m=(l+r)>>1;
    build(lson);
    build(rson);
}
int query(int x,int l,int r,int rt)
{
    if(l==r)
    {
        sum[rt]-=x;
        return l;
    }
    int m=(l+r)>>1;
    int ans=0;
    if(sum[rt<<1]>=x)
        ans=query(x,lson);
    else
        ans=query(x,rson);
    pushup(rt);
    return ans;
}
int main ()
{
    int x;
    while(scanf("%d%d%d",&h,&w,&n)!=EOF)
    {
        if(h>n)
            h=n;
        build(1,h,1);
        while(n--)
        {
            scanf("%d",&x);
            if(sum[1]<x)
                printf("-1
");
            else
                printf("%d
",query(x,1,h,1));
        }
    }
}

成段更新：

第一次写，学到了一个叫做懒惰标记或者叫做延时标记的东西，所谓的延时标记就是更新的时候不是更新到最底层的叶子节点，而是当覆盖整个子区间的时候做一个标记，当有一次update或者query到这个区间的时候把标记传递下去。

hdu1698 Just a Hook

Problem Description

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.

 

Source

2008 “Sunline Cup” National Invitational Contest

题目大意：

定义操作 i j  x把i~j区间的奖牌颜色变为x(1/2/3)

给你n个数(1~n)以及Q个操作最后求1~n 的奖牌加权和(权值就是颜色编号)

分析：

线段树 区间替换+求和

利用上面说的延时标记更新结点，区间求和跟上面的一样，这里只有一次询问且询问区间为[1,n]，所以更新完后最终答案就是根结点的sum[1]。

#include <iostream>
#include <cstdio>
#include <cstring>
#define maxlen 100010
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
int sum[maxlen<<2];
int col[maxlen<<2];
void pushup(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void pushdown(int rt,int m)
{
    if(col[rt])
    {
        col[rt<<1]=col[rt<<1|1]=col[rt];
        sum[rt<<1]=(m-(m>>1))*col[rt];
        sum[rt<<1|1]=(m>>1)*col[rt];
        col[rt]=0;
    }
}
void build(int l,int  r,int rt)
{
    sum[rt]=1;
    col[rt]=0;
    if(l==r)
        return ;
    int m=(l+r)>>1;
    build(lson);
    build(rson);
}
void update(int L,int R,int c,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        col[rt]=c;
        sum[rt]=c*(r-l+1);
        return ;
    }
    pushdown(rt,r-l+1);
    int m=(l+r)>>1;
    if(L<=m)
        update(L,R,c,lson);
    if(R>m)
        update(L,R,c,rson);
    pushup(rt);
}
int main ()
{
    int t,n,m;
    scanf("%d",&t);
    for(int Case=1;Case<=t;++Case)
    {
        scanf("%d%d",&n,&m);
        build(1,n,1);
        while(m--)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            update(a,b,c,1,n,1);
        }
        printf("Case %d: The total value of the hook is %d.
",Case,sum[1]);
    }
}

poj3468 A Simple Problem with Integers

Description

You have N integers, A₁, A₂, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

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

题目大意：

1、Q i  j   sum[i]+sum[i+1]+...+sum[j]

2、C i j c  sum[k]+=c   i<=k<=j

分析：跟上面的类似，成段的加减+区间求和

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define maxlen 100010
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
long long sum[maxlen<<2];
long long add[maxlen<<2];
void pushup(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void pushdown(int rt,int m)
{
    if(add[rt])
    {
        add[rt<<1]+=add[rt];
        add[rt<<1|1]+=add[rt];
        sum[rt<<1]+=add[rt]*(m-(m>>1));
        sum[rt<<1|1]+=add[rt]*(m>>1);
        add[rt]=0;
    }
}
void build(int l,int r,int rt)
{
    add[rt]=0;
    if(l==r)
    {
        scanf("%I64d",&sum[rt]);
        return ;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    pushup(rt);
}
void update(int L,int R,int c,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        add[rt]+=c;
        sum[rt]+=c*(r-l+1);
        return ;
    }
    pushdown(rt,r-l+1);
    int m=(l+r)>>1;
    if(L<=m)
        update(L,R,c,lson);
    if(m<R)
        update(L,R,c,rson);
    pushup(rt);
}
long long  query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)
        return sum[rt];
    pushdown(rt,r-l+1);
    int m=(l+r)>>1;
    long long ans=0;
    if(L<=m)
        ans+=query(L,R,lson);
    if(m<R)
        ans+=query(L,R,rson);
    return ans;
}
int main ()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        build(1,n,1);
        while(m--)
        {
            char cmd[10];
            int a,b,c;
            scanf("%s",cmd);
            if(cmd[0]=='Q')
            {
                scanf("%d%d",&a,&b);
                printf("%I64d
",query(a,b,1,n,1));
            }
            else if(cmd[0]=='C')
            {
                scanf("%d%d%d",&a,&b,&c);
                update(a,b,c,1,n,1);
            }
        }
    }
}

 

poj3667 Hotel

题意:

1 a:询问是不是有连续长度为a的空房间,有的话住进最左边
2 a b:将[a,a+b-1]的房间清空
思路:记录区间中最长的空房间
线段树操作:update:区间替换 query:询问满足条件的最左断点

msum[i]表示以根结点编号为i的剩余最大房间数

lsum[i]表示前缀最大房间数

rsum[i]表示后缀最大房间数

col[i]为染色标记，-1表示初始，1表示有人住，0表示没有人主。

主要就区间合并的问题，将左半边的与右半边的合并

如果rsum[left son]+lsum[right son]>=d 那么就说明存在这样一个横跨两棵子树的区间，由于我们知道左子树最右边点的坐标x，那么这个连续空区间的第一个位置自然就是x-rsum[left son]+1。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define maxlen 60010
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
int msum[maxlen<<2],lsum[maxlen<<2],rsum[maxlen<<2],col[maxlen<<2];
int max(int a,int b)
{
    return a>b?a:b;
}
void pushdown(int rt,int m)
{
    if(col[rt]!=-1)
    {
        col[rt<<1]=col[rt<<1|1]=col[rt];
        msum[rt<<1]=lsum[rt<<1]=rsum[rt<<1]=col[rt]?0:m-(m>>1);
        msum[rt<<1|1]=lsum[rt<<1|1]=rsum[rt<<1|1]=col[rt]?0:(m>>1);
        col[rt]=-1;
    }
}
void pushup(int rt,int m)
{
    lsum[rt]=lsum[rt<<1];
    rsum[rt]=rsum[rt<<1|1];
    if(lsum[rt]==m-(m>>1))
        lsum[rt]+=lsum[rt<<1|1];
    if(rsum[rt]==(m>>1))
        rsum[rt]+=rsum[rt<<1];
    msum[rt]=max(lsum[rt<<1|1]+rsum[rt<<1],max(msum[rt<<1],msum[rt<<1|1]));
}

void build(int l,int r,int rt)
{
    msum[rt]=lsum[rt]=rsum[rt]=r-l+1;
    col[rt]=-1;
    if(l==r)
        return ;
    int m=(l+r)>>1;
    build(lson);
    build(rson);
}
void update(int L,int R,int c,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        msum[rt]=lsum[rt]=rsum[rt]=c?0:(r-l+1);
        col[rt]=c;
        return ;
    }
    pushdown(rt,r-l+1);
    int m=(l+r)>>1;
    if(L<=m)
        update(L,R,c,lson);
    if(m<R)
        update(L,R,c,rson);
    pushup(rt,r-l+1);
}
int query(int sum,int l,int r,int rt)
{
    if(l==r)
        return l;
    pushdown(rt,r-l+1);
    int m=(l+r)>>1;
    if(msum[rt<<1]>=sum)
        return query(sum,lson);
    else if(rsum[rt<<1]+lsum[rt<<1|1]>=sum)
        return m-rsum[rt<<1]+1;
    return query(sum,rson);
}
int main ()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int cmd,x,y;
        build(1,n,1);
        while(m--)
        {
            scanf("%d",&cmd);
            if(cmd==1)
            {
                scanf("%d",&x);
                if(msum[1]<x)
                    printf("0
");
                else
                {
                    int ans=query(x,1,n,1);
                    printf("%d
",ans);
                    update(ans,ans+x-1,1,1,n,1);
                }
            }
            else
            {
                scanf("%d%d",&x,&y);
                update(x,x+y-1,0,1,n,1);
            }
        }
    }
}

未完待续...