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

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+, M = 1e6+, mod = 1e6+, inf = ;

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 = ;

        pos[] = ;

        memset(b,,sizeof(b));

        memset(sum,,sizeof(sum));

        memset(add,,sizeof(add));

        for(int i = ; i <= n; ++i) b[i] = a[i];

        for(int i = ; i <= n; ++i) sum[i] = sum[i-] + b[i];

        for(int i = ; i < N; ++i) mi[i] = inf;

        for(int i = ; i <= n; ++i) {

            if(pos[i] != pos[i-]) s = ;

            s += b[i];

            mi[pos[i]] = min(mi[pos[i]],s);

        }

        for(int i = ; i <= n; ++i) add[pos[i]] += b[i];

        f[] = ;

        s = ;

        for(int i = ; i <= n; ++i) {

            s+=b[i];

            if(s < ) f[i] = ;

            else f[i] = f[i-];

        }

}

int solve(int l,int r) {

        int OK = ;

        swap(b[l],b[r]);

        if(pos[l] == pos[r]) {

                int s = sum[block * (pos[l]-)];

                for(int i = block * (pos[l]-) + ; i <= min(pos[l] * block,n); ++i) {

                    s += b[i];

                    if(s < ) OK = ;

                }

                for(int i = pos[l]+; i <= m; ++i) {

                    if(s + mi[i] < ) OK = ;

                    s += (add[i]);

                }if(s < ) OK = ;

        } else {

                int s = sum[block * (pos[l]-)];

                for(int i = block*(pos[l]-) + ; i <= min(pos[l] * block,n); ++i) {

                    s += b[i];

                    if(s < ) OK = ;

                }

                for(int i = pos[l]+; i <= pos[r]-; ++i) {

                    if(s + mi[i] < ) OK = ;

                    s += (add[i]);

                }

                for(int i = block * (pos[r]-) + ; i <= min(pos[r]*block,n); ++i) {

                    s += b[i];

                    if(s < ) OK = ;

                }

                  for(int i = pos[r]+; i <= m; ++i) {

                    if(s + mi[i] < ) OK = ;

                    s += (add[i]);

                }

        }

        swap(b[l],b[r]);

        if(OK == ) return ;

        else return ;

}

int main() {

        while(scanf("%d%d",&n,&q)!=EOF) {

            scanf("%s",ch);

            block = int(sqrt(n));

            memset(a,,sizeof(a));

            memset(pos,,sizeof(pos));

            for(int i = ; i <= n; i++) {if(ch[i-] == '(') a[i] = ; else a[i] = -; pos[i]=(i-)/block+;}

            if(n%block) m = n/block + ; else m = n/block;

            init();

            for(int i = ; i <= q; ++i) {

                int l,r;

                scanf("%d%d",&l,&r);

                if(l > r) swap(l,r);

                if(solve(l,r) == ) printf("Yes\n");

                else printf("No\n");

            }

        }

        return ;

}