Parenthesis（括号平衡串，匹配问题）

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

题意：给定一个长度为n的括号平衡串，q次交换，问交换后是不还是平衡串？ 是输出Yes,不是输出No。

分析： 未交换时，就是一个平衡串，即括号都能够匹配
那么 交换的位置 可能有以下几种情况：
1.原来就是一样的 或者 左边的是’)’，右边的是’(’ 直接特判 （少一种判断会TLE）
2.左边的是’(’ ,右边的是’)’ 计数判断

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int mod=10056;const int N=1e5+5;typedef long long LL;#define mem(a,n) memset(a,n,sizeof(a))char str[N];int main(){    int n,q;    while(~scanf("%d%d",&n,&q))    {        scanf("%s",str);        for(int i=0; i<q; i++)        {            int a,b;            scanf("%d%d",&a,&b);            if(str[a-1]==str[b-1])            {                puts("Yes");                continue;            }            if(a>b) swap(a,b);            if(str[a-1]==')'&&str[b-1]=='(')            {                puts("Yes");                continue;            }            swap(str[a-1],str[b-1]);            int ans=0;            for(int i=0; i<n; i++)            {                if(str[i]=='(')                    ans++;                else ans--;                if(ans<0) break;            }            printf("%s\n",ans?"No":"Yes");            swap(str[a-1],str[b-1]);///  还要换回去！        }    }    return 0;}