Parenthesis(前缀和+线段树)

1809: Parenthesis

Time Limit: 5 Sec     Memory Limit: 128 Mb     Submitted: 2291     Solved: 622

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Description

Bobo has a balanced parenthesis sequence P=p1 p2…pn of length n and q questions.

The i-th question is whether P remains balanced after pai and pbi  swapped. Note that questions are individual so that they have no affect on others.

Parenthesis sequence S is balanced if and only if:

1. S is empty;

2. or there exists balanced parenthesis sequence A,B such that S=AB;

3. or there exists balanced parenthesis sequence S' such that S=(S').

Input

The input contains at most 30 sets. For each set:

The first line contains two integers n,q (2≤n≤105,1≤q≤105).

The second line contains n characters p1 p2…pn.

The i-th of the last q lines contains 2 integers ai,bi (1≤ai,bi≤n,ai≠bi).

Output

For each question, output "Yes" if P remains balanced, or "No" otherwise.

Sample Input

4 2
(())
1 3
2 3
2 1
()
1 2

Sample Output

No
Yes
No

Hint

Source

湖南省第十二届大学生计算机程序设计竞赛

//题意: n 长字符串，m次询问，一开始括号是匹配的，问 a ，b 位置的字符调换后，是否匹配

//题解:网上很多的暴力超时代码没看懂，这题，如果，( 看成 1 ，) 看成 -1 ，其实就是求 a -- (b-1) 中前缀和最小的，在调换后是否能满足要求即可，写了个线段树求区间最小

 # include <cstdio>
 # include <cstring>
 # include <cstdlib>
 # include <iostream>
 # include <vector>
 # include <queue>
 # include <stack>
 # include <map>
 # include <bitset>
 # include <sstream>
 # include <set>
 # include <cmath>
 # include <algorithm>
 #pragma comment(linker,"/STACK:102400000,102400000")
 using namespace std;
 #define LL          long long
 #define lowbit(x)   ((x)&(-x))
 #define PI          acos(-1.0)
 #define INF         0x3f3f3f3f3f3f3f3f
 #define eps         1e-8
 #define MOD         1000000007

 inline int scan() {
     int x=,f=; char ch=getchar();
     while(ch<''||ch>''){if(ch=='-') f=-; ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-''; ch=getchar();}
     return x*f;
 }
 inline void Out(int a) {
     if(a<) {putchar('-'); a=-a;}
     if(a>=) Out(a/);
     putchar(a%+'');
 }
 #define MX 100005
 /**************************/
 struct Tree
 {
     int l,r;
     int m;
 }tree[MX*];

 int st[MX];
 char str[MX];

 void build_tree(int l,int r,int k)
 {
     tree[k] = (Tree){l,r,};
     if (l==r)
     {
         tree[k].m = st[l];
         return;
     }
     int mid = (l+r)/;
     build_tree(l,mid,*k);
     build_tree(mid+,r,*k+);
     tree[k].m = min(tree[*k].m,tree[*k+].m);
 }

 int inqy(int l,int r,int k)
 {
     if (l==tree[k].l&&r==tree[k].r)
         return tree[k].m;

     int mid = (tree[k].l+tree[k].r)/;
     if (r<=mid) return inqy(l,r,*k);
     else if (l>mid) return inqy(l,r,*k+);
     return min (inqy(l,mid,*k) , inqy(mid+,r,*k+));
 }

 int main()
 {
     int n,m;
     while (scanf("%d%d",&n,&m)!=EOF)
     {
         scanf("%s",str+);
         st[]=;
         for (int i=;i<=n;i++)
             st[i] = st[i-] + (str[i]=='('?:- );
         build_tree(,n,);
         while (m--)
         {
             int a = scan();
             int b = scan();
             if (a>b) swap(a,b);

             int ok=;

             int sb = inqy(a,b-,);
             if (str[a]=='(') sb--;
             else sb++;
             if (str[b]==')') sb--;
             else sb++;
             if (sb<) ok=;

             if (ok)
                 printf("Yes\n");
             else
                 printf("No\n");
         }
     }
     return ;
 }