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).
OutputFor each question, output "
Yes" if P remains balanced, or "
No" otherwise.Sample Input
4 2 (()) 1 3 2 3 2 1 () 1 2Sample Output
No Yes NoHint
是真的被自己菜到 了好烦看到括号就不想做
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1e5+10;
char str [maxn];
int main()
{
int n,q,x,y,a,b;
while(scanf("%d%d",&n,&q)!=EOF)
{
scanf("%s",str+1);
while(q--)
{
scanf("%d%d",&a,&b);
x = min(a,b);
y = max(a,b);
if(str[x] == str[y])
{
printf("Yes\n");
continue;
}else if(str[x]==')'&&str[y]=='(')
{
printf("Yes\n");
continue;
}else
{
swap(str[x],str[y]);
int sum = 0;
for(int i =1;i<=n;i++)
{
if(str[i]=='(')sum++;
else sum--;
if(sum<0) break;
}
if(sum==0)printf("Yes\n");
else printf("No\n");
swap(str[x],str[y]);
}
}
}
return 0 ;
}