湖南省第十二届大学生计算机程序设计竞赛 Parenthesis

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

想法：

关键读懂题意

根据合法括号的定义，(总是要比)多或者相等，就是说，前缀和必须总是>=0（对于每一个位置必须都要>=0， 最后一个必须等于0。

代码：

#include<stdio.h>
#include<string.h>
#define M 100001
int a[M],b[M];
char s[M];
int i,j,k,n,m,q;
void swap(int &a,int &b)
{
    int t;
    t=a;
    a=b;
    b=t;
}
int main()
{
    while(~scanf("%d%d",&n,&q))
    {
        scanf("%s",s);
        for(i=0;i<n;i++)
        {
            if(s[i]=='(')a[i+1]=1;
            else a[i+1]=-1;
            b[i+1]=b[i]+a[i+1];
        }
        for(i=0;i<q;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            if(a[x]==a[y])
            {
                printf("Yes\n");
                continue;
            }
            if(x>y)//让x<y
            swap(x,y);
            bool flag=1;
            int ans=b[x-1];
            ans+=a[y];
            if(ans<0)flag=0;//换后，到原x位置是否平衡
            else
            {
            for(j=x+1;j<y;j++)//换后，原x+1到原y-1位置中所有位置上是否平衡
            {
                ans+=a[j];
                if(ans<0){flag=0;break;}
            }
            ans+=a[x];
            if(ans<0)flag=0;//到原y位置上是否平衡
            }
            if(flag)printf("Yes\n");
            else printf("No\n");


        }
    }
    return 0;
}