原题链接

题解

首先，假设当前 \(s(l,r)\) 括号序列为合法序列，则有如下几种情况：

• \(l+1==r\) ()

• \(match[r]==l\) (...)

• \(match[r]!=l\) (...)...(...)

code

#include<bits/stdc++.h>
using namespace std;

const long long MOD = 1e9+7;

long long dp[705][705][3][3] = {0};
int match[705];

void dfs(int l, int r) {
    if (l + 1 == r) {
        dp[l][r][0][1] = dp[l][r][1][0] = dp[l][r][0][2] = dp[l][r][2][0] = 1;
        return;
    }

    if (match[r] == l) {
        dfs(l + 1, r - 1);

        for (int i = 0; i <= 2; i++) {
            for (int j = 0; j <= 2; j++) {
                if (j != 1) dp[l][r][0][1] = (dp[l][r][0][1] + dp[l + 1][r - 1][i][j]) % MOD;
                if (i != 1) dp[l][r][1][0] = (dp[l][r][1][0] + dp[l + 1][r - 1][i][j]) % MOD;
                if (j != 2) dp[l][r][0][2] = (dp[l][r][0][2] + dp[l + 1][r - 1][i][j]) % MOD;
                if (i != 2) dp[l][r][2][0] = (dp[l][r][2][0] + dp[l + 1][r - 1][i][j]) % MOD;
            }
        }
        return;
    }

    int r1 = match[r] - 1, l1 = match[r];
    dfs(l, r1);
    dfs(l1, r);

    for (int i = 0; i <= 2; i++) {
        for (int j = 0; j <= 2; j++) {
            for (int k = 0; k <= 2; k++) {
                for (int m = 0; m <= 2; m++)
                {
                    if(j==1&&k==1||j==2&&k==2) continue;
                    //if(i==j||i&&j) continue;if(k==m||k&&m) continue;这个要注释掉是因为另外半部分的外围括号可能不是配对的
                    dp[l][r][i][m] = (dp[l][r][i][m] + dp[l][r1][i][j] * dp[l1][r][k][m] % MOD) % MOD;
                }
            }
        }
    }
}

void solve() {
    string s;
    cin >> s;
    int n = s.size();
    stack<int> q;

    for (int i = 0; i < n; i++) {
        if (s[i] == ')') {
            match[i] = q.top();
            q.pop();
        } else {
            q.push(i);
        }
    }

    dfs(0, n - 1);


    long long ans = 0;
    for (int i = 0; i <= 2; i++) {
        for (int j = 0; j <= 2; j++) {
            ans = (ans + dp[0][n - 1][i][j]) % MOD;
        }
    }

    cout << ans << '\n';
}

signed main() {
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int TT = 1;
    // cin >> TT;
    while (TT--) solve();
    return 0;
}