2016年省赛G题， Parenthesis

Problem G: Parenthesis

Time Limit: 5 Sec  Memory Limit: 128 MB
Submit: 398  Solved: 75
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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

比赛的时候，老O查个单词，我后来又查了一下，单词没看完意思，说是"不规则的"，发现里面还有一个括号的意思，也就是说只有括号，当时还以为有其他字符，想了好久，好像其他字符在这里没有起到作用
然后是第二个条件，可不可以反推，纠结了好久，最后还是直接模拟了一遍，虽然已经知道会tle，还是写了一下，当时只有这个题目A得人比较多一点。

解题报告：
1、只有括号
2、3个条件就是括号匹配的实质。

思路：
　　直接栈模拟是肯定不行的，就算要用栈模拟也不是普通的栈模拟，直接记录左括号的个数，这样模拟。然后这里询问次数10^5，n = 10^5，就是10^10了，这里用线段树优化，每次)(区间都是+2,
　　直接yes,要是()，就看-2是否是小于0.

#include <stdio.h>
#include <algorithm>

using namespace std;
int temp[100005];
char str[100005];

#define INF 0x3f3f3f3f

struct node
{
    int left,right,val;
} c[100005*4];

void build_tree(int l,int r,int root)
{
    c[root].left=l;
    c[root].right=r;
    if(l==r)
    {
        c[root].val=temp[l];
        return ;
    }
    int mid=(l+r)/2;
    build_tree(l,mid,root*2);
    build_tree(mid+1,r,root*2+1);
    c[root].val=min(c[root*2].val,c[root*2+1].val);
}

void query(int l,int r,int &min1,int root)
{
    if(c[root].left==l&&c[root].right==r)
    {
        min1=c[root].val;
        return ;
    }
    int mid=(c[root].left+c[root].right)/2;
    if(mid<l)
        query(l,r,min1,root*2+1);
    else if(mid>=r)
        query(l,r,min1,root*2);
    else
    {
        int min2;
        query(l,mid,min1,root*2);
        query(mid+1,r,min2,root*2+1);
        min1=min(min1,min2);
    }
}

int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        scanf("%s",str+1);
        int counts = 0;
        for(int i=1; i<=n; i++)
        {
            if(str[i]=='(')
                temp[i] = ++counts;
            else temp[i] = --counts;
        }

        build_tree(1,n,1);
        int ls,rs;
        for(int i=0; i<m; i++)
        {
            scanf("%d%d",&ls,&rs);
            if(ls > rs)
                swap(ls,rs);

            if(str[ls]==')'&&str[rs]=='(')
            {
                puts("Yes");
                continue;
            }
            if(str[ls]==str[rs])
            {
                puts("Yes");
                continue;
            }
            if(str[ls]=='('&&str[rs]==')')
            {
                int mins = INF;
                query(ls,rs-1,mins,1);
                if(mins<2)
                    puts("No");
                else puts("Yes");
                continue;
            }
        }
    }
    return 0;
}