2016年湖南省第十二届大学生计算机程序设计竞赛：G—parathesis

题目链接：传送门

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 32 32 1()1 2

Sample Output

NoYesNo

解题思路：前缀和+RMQ

用a存储‘（’+1，‘）’-1

若满足匹配，则有任意a[i]>=0

交换两个位置的括弧：‘（’和‘（’，‘）’和‘）’，‘）’和‘（’都不会影响序列的匹配

只有‘（’和‘）’的情况下且a[i]到a[j-1]的最小值小于2才会使序列不匹配

#include <cstdio>  #include <cstring>  #include <cmath>  #include <iostream>  #include <queue>#include <set>#include <string>#include <stack>#include <algorithm>#include <map>using namespace std;  typedef unsigned long long ll;const int N = 100100;const int M = 20;const int mod = 1e9+7;const int INF = 0x3fffffff;int dp[N][M];int a[N];void RMQ( int  n ){for( int i = 1 ; i <= n ; ++i ) dp[i][0] = a[i];for( int j = 1 ; (1<<j) <= n ; ++j ){for( int i = 1 ; i+(1<<j)-1 <= n ; ++i ){dp[i][j] = min( dp[i][j-1] , dp[i+(1<<(j-1))][j-1] );}}}int solve( int l , int r ){int k = 0;while( 1<<(k+1)<=r-l+1 ) k++;return min( dp[l][k] , dp[r-(1<<k)+1][k] );}int main(){int n,q,t1,t2;while( ~scanf("%d%d",&n,&q) ){memset( a , 0 , sizeof(a) );string b;cin >> b;int sz = b.size();a[0] = 0;for( int i = 1 ; i <= sz ; ++i ){if( b[i-1] == '(' ) a[i] = a[i]+a[i-1]+1;else a[i] = a[i]+a[i-1]-1;}RMQ( sz );for( int i = 0 ; i < q ; ++i ){int flag = 0;scanf("%d%d",&t1,&t2);if( t1 > t2 ) swap( t1 , t2 );if( b[t1-1] == '(' && b[t2-1] == ')' ){int t = solve(t1,t2-1);if( t<2 ) flag = 1;}if( flag ) printf("No\n");else printf("Yes\n");}}return 0;}