ACM2016湖南省赛Parenthesis_parenthesisacm-优快云博客

本文链接：https://blog.youkuaiyun.com/xky140610205/article/details/52432529

本文介绍了一种使用线段树解决平衡括号序列问题的方法。该问题要求判断交换两个位置的括号后，序列是否仍保持平衡。文章通过具体实例展示了如何构建线段树，并在查询过程中快速判断括号序列的有效性。

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1809: Parenthesis

Time Limit: 5 Sec Memory Limit: 128 MB
Submit: 589 Solved: 142
[ Submit][ Status][ Web Board]

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

HINT

Source

湖南省第十二届大学生计算机程序设计竞赛

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1809

湖南省赛心都碎了

给你一个(平衡)的一些括号问你随便调换两个位置的括号能不能再平衡

刚开始暴力了每次调换检验一次遇到 ( ++ )-- 一旦-1打断还是TE了

所以觉得有规律比赛时找了几个小时规律还是没找到

竟然是线段树省时间啦还是找到一点规律

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define LL 100010
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
int tree[LL*4],ans[LL];
int flag;
void pushup(int rt)
{
    if(tree[rt<<1|1]==1||tree[rt<<1]==1)
        tree[rt]=1;
}
void build(int l,int r,int rt)
{
    tree[rt]=0;
    if(l==r)
    {
        if(ans[l-1]<2)
            tree[rt]=1;
        return ;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    pushup(rt);
}

void query(int L,int R,int l,int r,int rt)
{
    if(flag) return ;
    if(L<=l&&R>=r)
    {
        if(tree[rt])
        {
            flag=1;
        }
        return ;
    }
    int m=(l+r)>>1;
    if(L<=m) query(L,R,lson);
    if(R>m)  query(L,R,rson);
}

int main()
{
    char str[100010];
    int a,b,n,m,i;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        scanf("%s",str);
            ans[0]=1;
        for(i=1; i<n; i++)
            if(str[i]=='(')
                ans[i]=ans[i-1]+1;
            else ans[i]=ans[i-1]-1;
        build(1,n,1);
        for(int i=0; i<m; i++)
        {
            scanf("%d%d",&a,&b);
            if(a>b)
                swap(a,b);
            if(str[a-1]==str[b-1])
            {
                printf("Yes\n");
                continue;
            }
            if(str[a-1]==')') //这样必然可以
            {
                printf("Yes\n");
                continue;
            }
            flag=0;
            query(a,b-1,1,n,1);//查询区间里有没有<2的
            if(!flag) printf("Yes\n");//如果有 把)换到前面去一定会导致这个区间都减掉2
            else printf("No\n");
        }
    }
    return 0;
}