CodeForces

Description

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

Solution

我已经菜到做 div.3 了

(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;
}