湖南省第十二届大学生计算机程序设计竞赛---Parenthesis（线段树求区间最值）

Posted 2020-08-06 茶飘香~

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

湖南省第十二届大学生计算机程序设计竞赛

 

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
*/