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想法：关键读懂题意根据合法括号的定义，(总是要比)多或者相等，就是说，前缀和必须总是>=0（对于每一个位置必须都要>=0， 最后一个必须等于0。代码：#include<stdio.h>#include<string.h>
#define M 100001
int a[M],b[M];
char s[M];
int i,j,k,n,m,q;
void swap(int &a,int &b)
{
int t;
t=a;
a=b;
b=t;
}
int main()
{
while(~scanf("%d%d",&n,&q))
{
scanf("%s",s);
for(i=0;i<n;i++)
{
if(s[i]=='(')a[i+1]=1;
else a[i+1]=-1;
b[i+1]=b[i]+a[i+1];
}
for(i=0;i<q;i++)
{
int x,y;
scanf("%d%d",&x,&y);
if(a[x]==a[y])
{
printf("Yes\n");
continue;
}
if(x>y)//让x<y
swap(x,y);
bool flag=1;
int ans=b[x-1];
ans+=a[y];
if(ans<0)flag=0;//换后，到原x位置是否平衡
else
{
for(j=x+1;j<y;j++)//换后，原x+1到原y-1位置中所有位置上是否平衡
{
ans+=a[j];
if(ans<0){flag=0;break;}
}
ans+=a[x];
if(ans<0)flag=0;//到原y位置上是否平衡
}
if(flag)printf("Yes\n");
else printf("No\n");
}
}
return 0;
}