"Weilai Cup" 2022 Niuke summer multi school training camp 2 k.[link with bracket sequence i] bracket sequence DP

K.Link with Bracket Sequence I (DP)

Topic analysis

Given length is $n$ The bracket sequence of ( No guarantee of legality ), The length generated on this basis is $m$ The number of schemes in the bracket sequence .

set up $dp[i][j][k]$ Indicates that the number of parentheses inserted is $i$、 The number of parentheses in the original sequence used is $j$, There are more left parentheses than right parentheses $k$ When the number of programs . Then the final answer is $dp[m-n][n][0]$. Then consider how to design state transition ：

First enumerate the number of parentheses inserted , The original parenthesis sequence and the number of left parentheses is more than that of right parentheses .

If the parentheses enumerated at present are left parentheses , And the number of original brackets used $<n$, You can put this bracket in the final sequence , That is to say ：
$$dp[i][j+1][k+1]=(dp[i][j+1][k+1]+dp[i][j][k])\%mod$$
If a right parenthesis is enumerated at this time , also $k>0$, That is, the number of left parentheses is greater than the number of right parentheses , And the number of original brackets used $<n$, Put the right parenthesis in the final sequence ：
$$dp[i][j+1][k-1]=(dp[i][j+1][k-1]+dp[i][j][k])\%mod$$
If the number of parentheses used is $n$, Or the current enumeration to the right bracket , The left parenthesis can be inserted ：
$$dp[i+1][j][k+1]=(dp[i+1][j][k+1]+dp[i][j][k])\%mod$$
When $k>0$ when , If the number of parentheses in the original sequence is $n$, Or the current enumeration to the left bracket , You can insert the right parenthesis ：
$$dp[i+1][j][k-1]=(dp[i+1][j][k-1]+dp[i][j][k])\%mod$$

Code

#include <bits/stdc++.h>
#pragma gcc optimize(2)
#pragma g++ optimize(2)
// #define int long long
#define endl '\n'
using namespace std;

const int N = 2e5 + 10, mod = 1e9 + 7;
int dp[220][220][220];

inline void solve(){
    
    // memset(dp, 0, sizeof(dp));
    int m, n; cin >> n >> m;
    string s; cin >> s;
    // if(n & 1 || (m - n) & 1) { cout << 0 << endl; return; }
    dp[0][0][0]=1;
	for (int i = 0; i <= m - n; ++i)
    for (int j = 0; j <= n; ++j)
        for (int k = 0; k <= m; ++k){
    
            if (s[j] == '(' && j < n)
                dp[i][j + 1][k + 1] = (dp[i][j + 1][k + 1] + dp[i][j][k]) % mod;
            else
                dp[i + 1][j][k + 1] = (dp[i + 1][j][k + 1] + dp[i][j][k]) % mod;
            if (k > 0){
    
                if (s[j] == ')' && j < n)
                    dp[i][j + 1][k - 1] = (dp[i][j + 1][k - 1] + dp[i][j][k]) % mod;
                else
                    dp[i + 1][j][k - 1] = (dp[i + 1][j][k - 1] + dp[i][j][k]) % mod;
            }
        }
	cout << dp[m - n][n][0] << endl;
    for(int i = 0; i <= m + 2; i++)
        for(int j = 0; j <= n + 2; j++)
            for(int k = 0; k <= m + 2; k++) dp[i][j][k] = 0;
}

signed main(){
    
    ios_base::sync_with_stdio(false), cin.tie(0);
    int t = 1; cin >> t;
    while(t--) solve();
    return 0;
}