第十二届湖南省赛G - Parenthesis (树状数组维护)

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).

 

OutputFor 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:

给你一个长度为N个合法的括号字符串,然后有 Q 个询问,每一个询问Q,有一个L和R,如果字符串中的L和R位置的两个字符交换后,括号字符串仍然合法的话,那么输出Yes,否则输出No。

 

思路:

可以用树状数组维护一下,

我们定义如下,如果字符是'(' 我们定他的权值为-1,')' 定义权值为 +1,

我们容易知道,一个合法的字符串的总权值和是0.,并且一个合法的括号串,不存在任意一个位置i,1~i到sum和不大于0。

因为题目给的是一个合法的字符串,那么每一次询问的时候,我们把对应的位置的数值给改变一下,

然后检测如下条件是否成立。

sum(1~l),sum( 1~l+1 ) ,sum(1~r),sum( 1~ r+1 )

需要以上的值均小于等于0,那么这个字符串一定是合法的。

我们可以通过树状数组来做,因为基础的树状数组模板就有单点修改,区间查询的功能。

细节见代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=100010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int tree[maxn];
int n,q;
char s[maxn];
int lowbit(int x)
{
    return x&(-1*x);
}
void add(int x,int k)
{
    for (int i = x; i <=n ; i+=lowbit(i))
    {
        tree[i]+=k;
    }
}
int query(int x){

    int res=0;
    while(x>0)
    {
        res+=tree[x];
        x-=lowbit(x);
    }
    return res;
}
int main()
{
//    freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
    //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
    while(~scanf("%d %d",&n,&q))
    {
        MS0(tree);
        scanf("%s",s+1);
        repd(i,1,n)
        {
            if(s[i]=='(')
            {
                add(i,-1);
            }else
            {
                add(i,1);
            }
        }
        int l,r;
        while(q--)
        {
            scanf("%d %d",&l,&r);
            if(s[l]==s[r])
            {
                printf("Yes\n");
            }else
            {
                if(s[l]=='(')
                {
                    // 改 为 )  -1 -> 1
                    add(l,2);
                    // 1  -> -1
                    add(r,-2);
                }else
                {
                    add(l,-2);
                    add(r,2);
                }
                if(query(l)<=0&&query(l+1)<=0&&query(r)<=0&&query(r+1)<=0&&query(n)==0)
                {
                    printf("Yes\n");
                }else
                {
                    printf("No\n");
                }
                if(s[l]=='(')
                {
                    add(l,-2);
                    add(r,2);
                }else
                {
                    add(l,2);
                    add(r,-2);
                }
            }
        }
    }




    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}

 

上一篇:leetcode -- 110. Balanced Binary Tree


下一篇:【CF1237C】Balanced Removals(降维)