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

Posted 2020-12-27 sssy

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;
}