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=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).

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

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   
#include   
#include   
#include   
#include 
#include 
#include 
#include 
#include 
#include 
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<> 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;
}






你可能感兴趣的:(数据结构---RMQ,前缀和)