CSU 1809: Parenthesis【前缀和】

1809: Parenthesis

Submit    Status    Time Limit: 5 Sec | Memory Limit: 128 Mb | Submitted: 1111 | Solved: 317

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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 32 32 1()1 2

Sample Output

NoYesNo

Hint

Source

湖南省第十二届大学生计算机程序设计竞赛

#include<iostream>#include<cstdio>#include<math.h>#include<cstring>#include<climits>#include<string>#include<queue>#include<stack>#include<set>#include<map>#include<list>#include<vector>#include<sstream>#include<algorithm>using namespace std;#define rep(i,j,k)for(i=j;i<k;i++)#define per(i,j,k)for(i=j;i>k;i--)#define ms(x,y)memset(x,y,sizeof(x))#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define ll long longconst int INF=0x7ffffff;const int M=1e5+10;int i,j,k,n,m,q;char s[M];int a[M];int b[M];int main(){    while(~scanf("%d%d",&n,&q)){        ms(a,0),ms(b,0);        getchar();        for(i=1;i<=n;i++){            scanf("%c",&s[i]);            if(s[i]=='(')a[i]=1;            else a[i]=-1;            b[i+1]=b[i]+a[i];        }        while(q--){            int x,y;            scanf("%d%d",&x,&y);            if(s[x]==s[y]){                printf("Yes\n");continue;            }            if(x>y)swap(x,y);            int flag=0;            for(i=x+1;i<=y;i++){                b[i]=b[i]-a[x]+a[y];                if(b[i]<0)flag=1;            }            if(b[y]+a[x]<0)flag=1;            for(i=x+1;i<=y;i++){                b[i]=b[i]+a[x]-a[y];            }            if(flag)printf("No\n");            else printf("Yes\n");        }    }    return 0;}