2016湖南省赛G题。

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终于还是写完了。。懒癌复发。。。原来可以用线段树。判断区间内最小的前缀是否小于2就行了。。。之前比赛的时候没有想到用这个方法。。。虽然自己会一点线段树。。但是可惜没写出来。。总的来说还是太菜了吧。要多刷刷线段树。。刷完线段树接下来就是最短路、最小生成树、DP了。

G - Parenthesis

Time Limit:5000MS Memory Limit:131072KB 64bit IO Format:%lld & %llu

Submit Status Practice CSU 1809

Appoint description:

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

#include<bits/stdc++.h>
using namespace std;
const int MAXN=1e5+5;
const int INF=0x3f3f3f3f;
int sum[MAXN<<2];
int pre[MAXN];
char s[MAXN];

void push_up(int k)
{
    sum[k]=min(sum[k<<1],sum[k<<1|1]);
}
void build(int l,int r,int k)
{
    if(l==r)
    {
        sum[k]=pre[l];
        return;
    }
    int mid=(l+r)>>1;
    build(l,mid,k<<1);
    build(mid+1,r,k<<1|1);
    push_up(k);
}

int query(int L,int R,int l,int r,int k)
{
    if(L<=l&&R>=r) return sum[k];
    int ans=INF;
    int mid=(l+r)>>1;
    if(L<=mid) ans=min(ans,query(L,R,l,mid,k<<1));
    if(R>mid) ans=min(ans,query(L,R,mid+1,r,k<<1|1));
    return ans;
}
int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m))
    {
        scanf("%s",s+1);
        for(int i=1;i<=n;i++)
        {
            if(s[i]=='(') pre[i]=pre[i-1]+1;
            else pre[i]=pre[i-1]-1;
        }
        build(1,n,1);
        while(m--)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            if(l>r)swap(l,r);
            if(s[l]==s[r]||(s[l]==')'&&s[r]=='(')) printf("Yes\n");
            else
            {
                if(query(l,r-1,1,n,1)<2) printf("No\n");
                else printf("Yes\n");
            }
        }
    }
    return 0;
}