原题链接
http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1809
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
Source
题意：给了一个平衡的括号序列s（平衡是指括号匹配正常） 现在q次询问，每次输入两个数a、b 问将s[a] s[b]交换后是否任然平衡，平衡则输出“Yes” 否则输出“No”；
思路：定义数组num[] ，num[i]表示s[1]到s[i]中左括号数减去右括号数的差值，分析可知因为s是平衡括号序列，那么num[i]>=0 令a<b ,那么交换s[a] s[b]后，只对num[a]~num[b-1]产生影响，并且交换后当num[k]<0（a<=k<b）时表示不平衡,而只有s[a]=‘(‘ s[b]=‘)‘ 交换后才可能使num[k]<0 。所以特判当s[a]=‘(‘ s[b]=‘)‘时，用线段树求区间a~b-1的最小值，当最小值小于2时，即交换后不平衡，为什么呢？ s[a]=‘(‘ s[b]=‘)‘交换后num[a]~num[b-1]都减2；
代码如下：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#define eps 1e-8
#define maxn 105
#define inf 0x3f3f3f3f3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;
char s[100005];
int num[100005];
int a,b;
struct Node
{
//int l,r;
int v;
}node[100005*4];
void build(int l,int r,int i)
{
if(l==r) {
node[i].v=num[l];
return;
}
int mid=(l+r)>>1;
build(l,mid,i<<1);
build(mid+1,r,i<<1|1);
node[i].v=min(node[i<<1].v,node[i<<1|1].v);
}
void query(int l,int r,int &tmp,int i)
{
if(l>=a&&r<=b) { tmp=node[i].v; return; }
int mid=(l+r)>>1;
if(mid>=b) query(l,mid,tmp,i<<1);
else if(mid<a) query(mid+1,r,tmp,i<<1|1);
else {
int tmp2;
query(l,mid,tmp,i<<1);
query(mid+1,r,tmp2,i<<1|1);
tmp=min(tmp,tmp2);
}
}
int main()
{
int n,q;
while(scanf("%d%d",&n,&q)!=EOF)
{
scanf("%s",s+1);
num[0]=0;
for(int i=1;i<=n;i++)
{
num[i]=num[i-1];
if(s[i]==‘(‘) num[i]++;
else num[i]--;
}
build(1,n,1);
while(q--)
{
scanf("%d%d",&a,&b);
if(a>b) swap(a,b);
if(s[a]==s[b]||s[a]==‘)‘&&s[b]==‘(‘) puts("Yes");
else
{
b--;
int tmp=99999999;
query(1,n,tmp,1);
if(tmp<2) puts("No");
else puts("Yes");
}
}
}
return 0;
}
/**
8 8
(())(())
2 7
*/