CodeForces - 1015F Bracket Substring

Description

给出一个括号序列 \(s(|s|\le 200)\) ，求长度为 \(2n(n\le 100)\) 且包含 \(s\) 作为字串的合法的括号序列个数。

Solution

\(dp[i][j][k]\) 表示现在正在填第 \(i\) 个字符，已经匹配的 \(s\) 长度为 \(j\) ，有 \(k\) 个未匹配的左括号。

转移显然，见代码。用 \(fail[]\) 加速即可。

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

template <class T> void read(T &x) {
    x = 0; bool flag = 0; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == 45) flag = 1;
    for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48; if (flag) x = -x;
}

#define N 205
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define P 1000000007

char s[N];
int fail[N], dp[N][N][N];

inline void calc(int& x, const int& y) { (x += y) >= P ? (x -= P) : 0; }

int main() {
    int n, m; read(n); n <<= 1;
    scanf("%s", s); m = strlen(s);
    rep(i, 1, m - 1) {
        int j = fail[i]; for (; j && s[i] != s[j]; j = fail[j]);
        fail[i + 1] = s[i] == s[j] ? j + 1 : 0;
    }
    dp[1][s[0] == '('][1] = 1;
    rep(i, 2, n) rep(j, 0, min(m, i)) rep(k, 0, i) {
        if (j < m) {
            if (s[j] == '(') calc(dp[i][j + 1][k + 1], dp[i - 1][j][k]);
            else {
                int nxt = j; for (; nxt && s[nxt] != '('; nxt = fail[nxt]);
                nxt += (s[nxt] == '(');
                calc(dp[i][nxt][k + 1], dp[i - 1][j][k]);
            }
            if (s[j] == ')') calc(dp[i][j + 1][k], dp[i - 1][j][k + 1]);
            else {
                int nxt = j; for (; nxt && s[nxt] != ')'; nxt = fail[nxt]);
                nxt += (s[nxt] == ')');
                calc(dp[i][nxt][k], dp[i - 1][j][k + 1]);
            }
        }
        else {
            calc(dp[i][j][k + 1], dp[i - 1][j][k]);
            calc(dp[i][j][k], dp[i - 1][j][k + 1]);
        }
    }
    cout << dp[n][m][0];
    return 0;
}