题目

4091. 括号序列

算法标签: 区间$dp$

思路

区间$dp$添加维表示形态$f[i][j][k]$, 对于每种形态考虑状态如何进行转移, 枚举的时候不能重复, 星号也要定义唯一的解析方式, 算法时间复杂度$O(n^3)$

代码

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

typedef long long LL;
const int N = 510, MOD = 1e9 + 7;

int n, m;
string s;
int f[N][N][5];

bool check(char a, char b) {
	return a == b || a == '?';
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);

	cin >> n >> m >> s;
	for (int len = 1; len <= n; ++len) {
		for (int i = 0; i + len - 1 < n; ++i) {
			int j = i + len - 1;
			if (len == 1) {
				if (check(s[i], '*')) f[i][j][2] = 1;
			}
			else {
				if (check(s[i], '(') && check(s[j], ')')) {
					if (len == 2) f[i][j][0] = 1;
					for (int k = 0; k < 5; ++k) f[i][j][0] = (f[i][j][0] + f[i + 1][j - 1][k]) % MOD;
				}
				if (len <= m && check(s[i], '*')) f[i][j][2] = f[i + 1][j][2];
				// 枚举分界点
				for (int k = i; k < j; ++k) {
					auto t = f[i][j], l = f[i][k], r = f[k + 1][j];
					t[1] = (t[1] + (l[0] + (LL) l[1] + l[3]) * r[0]) % MOD;
					t[3] = (t[3] + (l[0] + (LL) l[1]) * r[2]) % MOD;
					t[4] = (t[4] + l[2] * (r[0] + (LL) r[1])) % MOD;
				}
			}
		}
	}

	int ans = ((LL) f[0][n - 1][0] + f[0][n - 1][1]) % MOD;
	cout << ans << "\n";

	return 0;
}