题目链接

Codeforces Round #501 (Div. 3) F. Bracket Substring

题解

官方题解

http://codeforces.com/blog/entry/60949 ....看不懂

设dp[i][j][l]表示前i位，左括号-右括号=j，匹配到l了

状态转移，枚举下一个要填的括号，用next数组求状态的l,分别转移

代码

#include<bits/stdc++.h>

using namespace std;

const int maxn = 207;

const int mod = 1e9 + 7;

char s[maxn];

int next[maxn],dp[maxn][maxn][maxn],n;

void GetNext(int l) {

    next[0] = 0;

    for(int i = 1; i <= l; i++) {

        int p = next[i-1];

        while(p > 0 && s[p] != s[i]) p = next[p-1];

        next[i] = p + 1;

    }

}

int main() {

	scanf("%d%s",&n,s + 1);

	int len = strlen(s + 1);

	GetNext(len);

	dp[0][0][0] = 1;

	for(int i = 0;i <= 2 * n;++ i)

		for(int j = 0;j <= n; ++ j) {

	  		for(int l = 0;l < len;++ l) {

	  			int x = l + 1;

	  			while(x > 0 && s[x] != '(') x = next[x - 1];

	  			dp[i + 1][j + 1][x] = (dp[i + 1][j + 1][x] + dp[i][j][l]) % mod;

				x = l + 1;

				while(x > 0 && s[x] != ')') x = next[x - 1];

				if(j > 0)

				dp[i + 1][j - 1][x] = (dp[i + 1][j - 1][x] + dp[i][j][l]) % mod;

			}

			dp[i + 1][j + 1][len] =(dp[i + 1][j + 1][len] + dp[i][j][len]) % mod;

			if(j > 0) dp[i + 1][j - 1][len] = (dp[i][j][len] + dp[i + 1][j - 1][len]) % mod;

		}

	cout << dp[2 * n][0][len] << endl;

	return 0;

}

Codeforces Round #501 (Div. 3) F. Bracket Substring的相关教程结束。