HDU 5629 Clarke and tree dp+prufer序列

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int maxn = 55;
const int mod = 1e9 + 7;
typedef long long LL;

int n;
LL dp[maxn][maxn][maxn];
int arr[maxn];

LL C[maxn][maxn];
void pre() {
    C[0][0] = 1;
    for (int i = 1; i < maxn; i++) {
        C[i][0] = 1;
        for (int j = 1; j <= i; j++) {
            C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
        }
    }
}

void init() {
    memset(dp, 0, sizeof(dp));
}

int main() {
    pre();
    int tc;
    scanf("%d", &tc);
    while (tc--) {
        init();
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &arr[i]);
        }
        dp[0][0][0] = 1;
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= i; j++) {
                for (int k = 0; k <= n - 2; k++) {
                    //不用第i+1个数
                    dp[i + 1][j][k] = (dp[i + 1][j][k] + dp[i][j][k]) % mod;
                    //用第i+1个数
                    for (int l = 1; l <= arr[i + 1] && l - 1 + k <= n - 2; l++) {
                        dp[i + 1][j + 1][k + l - 1] = (dp[i + 1][j + 1][k + l - 1] + C[k + l - 1][l - 1] * dp[i][j][k]) % mod;
                    }
                }
            }
        }
        printf("%d", n);
        for (int i = 2; i <= n; i++) printf(" %lld", dp[n][i][i - 2]);
        printf("\n");
    }
    return 0;
}