Parenthesis

 

G - Parenthesis

Time Limit:5000MS     Memory Limit:131072KB     64bit IO Format:%lld & %llu

Description

Bobo has a balanced parenthesis sequence P=p ₁ p ₂…p _n of length n and q questions.

The i-th question is whether P remains balanced after p _ai and p _bi  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≤10 ⁵,1≤q≤10 ⁵).

The second line contains n characters p ₁ p ₂…p _n.

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

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

 思路:线段树；

 当括号匹配时我们有所有的前缀和sum〉=0;那么当我们交换两个括号时仅会改变sum[a]----sum[b]的值 ，并且只有当a为（  b为 ）才可能不匹配，那么当这两交换，这个区间段所有的值会减去2，那么我们用线段树维护sum的最小值即可，然后与2比较。复杂n*log(n)

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stdlib.h>
#include<string.h>
using namespace std;
char str[100005];
int tree[100005*4];
int ans[100005];
void build(int l,int r,int k);
int ask(int l,int r,int k,int nn,int mm);
int main(void)
{
    int n,m;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        scanf("%s",str);
        int l = strlen(str);
        ans[0] = 0;
        for(int i = 0; i < l; i++)
        {
            if(str[i] == '(')
            {
                ans[i+1] = 1;
            }
            else ans[i+1] = -1;
        }
        for(int i = 1; i <= l; i++)
        {
            ans[i] = ans[i] + ans[i-1];
        }
        build(1,l,0);
        while(m--)
        {
            int x,y;
            scanf("%d %d",&x,&y);
            if(x > y)swap(x,y);
            if(str[x-1]==str[y-1]||str[x-1]==')')
            {
                printf("Yes
");
            }
            else
            {
                int ak = ask(x,y-1,0,1,l);
                if(ak<2)
                    printf("No
");
                else printf("Yes
");
            }
        }
    }return 0;
}
void build(int l,int r,int k)
{
    if(l == r)
    {
        tree[k] = ans[l];
    }
    else
    {
        build(l,(l+r)/2,2*k+1);
        build((l+r)/2+1,r,2*k+2);
        tree[k] = min(tree[2*k+1],tree[2*k+2]);
    }
}
int ask(int l,int r,int k,int nn,int mm)
{
    if(nn>r||mm<l)
    {
      return 1e9;
    }
    else if(l<=nn&&r>=mm)
    {
        return tree[k];
    }
    else
    {
        int x = ask(l,r,2*k+1,nn,(nn+mm)/2);
        int y = ask(l,r,2*k+2,(nn+mm)/2+1,mm);
        return min(x,y);
    }
}