湖南省第十二届大学生计算机程序设计竞赛 G Parenthesis

1809: Parenthesis

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

HINT

 

题意：

　　给你长度n的合法括号匹配和q个询问

　　每次询问你 交换ai,bi两个位置的符号，交换后是否还是合法的

题解：

　　分块

　　（转化为1，）转化为-1

　　合法括号序列满足前缀和永远大于等于0

　　利用这个每次修改两个位置

　　维护这个关系就好了

　　分块可过的

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
 
typedef long long LL;
const long long INF = 1e18;
const double Pi = acos(-1.0);
const int N = 2e5+10, M = 1e6+11, mod = 1e6+3, inf = 5000;
 
int block,n,q,m;
int sum[N],pos[N],mi[N],a[N],b[N],add[N],f[N];
char ch[N];
void init() {
        int s = 0;
        pos[0] = 0;
        memset(b,0,sizeof(b));
        memset(sum,0,sizeof(sum));
        memset(add,0,sizeof(add));
        for(int i = 1; i <= n; ++i) b[i] = a[i];
        for(int i = 1; i <= n; ++i) sum[i] = sum[i-1] + b[i];
        for(int i = 0; i < N; ++i) mi[i] = inf;
        for(int i = 1; i <= n; ++i) {
            if(pos[i] != pos[i-1]) s = 0;
            s += b[i];
            mi[pos[i]] = min(mi[pos[i]],s);
        }
        for(int i = 1; i <= n; ++i) add[pos[i]] += b[i];
        f[0] = 1;
        s = 0;
        for(int i = 1; i <= n; ++i) {
            s+=b[i];
            if(s < 0) f[i] = 0;
            else f[i] = f[i-1];
        }
}
int solve(int l,int r) {
        int OK = 1;
        swap(b[l],b[r]);
        if(pos[l] == pos[r]) {
                int s = sum[block * (pos[l]-1)];
                for(int i = block * (pos[l]-1) + 1; i <= min(pos[l] * block,n); ++i) {
                    s += b[i];
                    if(s < 0) OK = 0;
                }
                for(int i = pos[l]+1; i <= m; ++i) {
                    if(s + mi[i] < 0) OK = 0;
                    s += (add[i]);
                }if(s < 0) OK = 0;
        } else {
                int s = sum[block * (pos[l]-1)];
                for(int i = block*(pos[l]-1) + 1; i <= min(pos[l] * block,n); ++i) {
                    s += b[i];
                    if(s < 0) OK = 0;
                }
                for(int i = pos[l]+1; i <= pos[r]-1; ++i) {
                    if(s + mi[i] < 0) OK = 0;
                    s += (add[i]);
                }
                for(int i = block * (pos[r]-1) + 1; i <= min(pos[r]*block,n); ++i) {
                    s += b[i];
                    if(s < 0) OK = 0;
                }
                  for(int i = pos[r]+1; i <= m; ++i) {
                    if(s + mi[i] < 0) OK = 0;
                    s += (add[i]);
                }
        }
        swap(b[l],b[r]);
        if(OK == 1) return 1;
        else return 0;
}
int main() {
        while(scanf("%d%d",&n,&q)!=EOF) {
            scanf("%s",ch);
            block = int(sqrt(n));
            memset(a,0,sizeof(a));
            memset(pos,0,sizeof(pos));
            for(int i = 1; i <= n; i++) {if(ch[i-1] == '(') a[i] = 1; else a[i] = -1; pos[i]=(i-1)/block+1;}
            if(n%block) m = n/block + 1; else m = n/block;
            init();
            for(int i = 1; i <= q; ++i) {
                int l,r;
                scanf("%d%d",&l,&r);
                if(l > r) swap(l,r);
                if(solve(l,r) == 1) printf("Yes\n");
                else printf("No\n");
            }
        }
        return 0;
}