CSU 1809: 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 3
2 3
2 1
()
1 2

Sample Output

No
Yes
No

Hint

Source

湖南省第十二届大学生计算机程序设计竞赛

代码：

#include<stdio.h>
#include<string.h>
int main()
{
int n,m;
char str[100010];
int s[100010];
int s1[100010];
while(scanf("%d%d",&n,&m)!=EOF)
{
int i,j;
scanf("%s",str);
int len=strlen(str);
s[0]=0;
s1[0]=0;
for(i=0;i<len;i++)
{
if(str[i]=='(')
{
s[i+1]=1;
s1[i+1]=s1[i]+1;
}
if(str[i]==')')
{
s[i+1]=-1;
s1[i+1]=s1[i]-1;
}
}
int a,b,t;
for(i=0;i<m;i++)
{
int sum=0;
int flag=0;
scanf("%d %d",&a,&b);
if(a>b)
{
t=a;a=b;b=t;
}
if(s[a]==s[b]||(s[a]==-1&&s[b]==1))
{
printf("Yes\n");
continue;
}
sum+=s1[a-1];
for(j=a;j<=b;j++)
{
if(sum<0)
{
flag=1;
break;
}
if(j==a||j==b)
{
if(j==a)
sum+=s[b];
else
sum+=s[a];
}
else  sum+=s[j];
}
sum+=s[len]-s[b];
if(sum<0)
{
flag=1;
}
if(flag) printf("No\n");
else
{
printf("Yes\n");
}
}
}
return 0;
}