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

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;
}
``````