删括号(dp)

题目链接：https://ac.nowcoder.com/acm/problem/21303

思路：删括号的时候一定要时刻保证左括号数量比右括号多，我们可以定义dp[i][j][k]表示考虑AA前i个匹配了B前j个A被删除部分左括号数-右括号数=k是否可行，

分类讨论转移即可，最后答案就是dp[n][m][0]。

#include <cstdio>
#include <map>
#include <iostream>
#include<cstring>
#include<bits/stdc++.h>
#define ll long long int
#define M 6
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
int dp[107][107][107];
int main(){
    ios::sync_with_stdio(false);
    string s,t;
    cin>>s>>t;
    int lens=s.length();
    int lent=t.length();
    dp[0][0][0]=1;
    for(int i=0;i<lens;i++)
        for(int j=0;j<=lent;j++)
            for(int k=0;k<=lens;k++){
                if(dp[i][j][k]){
                    if(!k&&s[i]==t[j]) dp[i+1][j+1][k]=1;
                    if(s[i]=='(') dp[i+1][j][k+1]=1;
                    else if(k) dp[i+1][j][k-1]=1;
                }
            }
    if(dp[lens][lent][0])
    cout<<"Possible"<<endl;
    else cout<<"Impossible"<<endl;
}