HDU-5378 概率DP

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=1e3+10;
const int P=1e9+7;
int n,k,son[N];
int dp[N][N];
vector<int> G[N];

LL power(LL x,LL p) {
    LL ret=1;
    for (;p;p>>=1) {
        if (p&1) ret=ret*x%P;
        x=x*x%P;
    }
    return ret;
}

void dfs(int x,int fa) {
    son[x]=1;
    for (int i=0;i<G[x].size();i++) {
        int y=G[x][i];
        if (y==fa) continue;
        dfs(y,x);
        son[x]+=son[y];
    }
}

int inv[N];
void prework() {
    for (int i=1;i<=1000;i++) inv[i]=power(i,P-2);
}

int main()
{
    prework();
    int T,cas=0; cin>>T;
    while (T--) {
        scanf("%d%d",&n,&k);
        for (int i=1;i<=n;i++) G[i].clear();
        for (int i=1;i<n;i++) {
            int x,y; scanf("%d%d",&x,&y);
            G[x].push_back(y);
            G[y].push_back(x);
        }
        for (int i=1;i<=n;i++) son[i]=0;
        dfs(1,0);
        
        for (int i=0;i<=n;i++) for (int j=0;j<=n;j++) dp[i][j]=0;
        dp[0][0]=1;
        for (int i=1;i<=n;i++) 
            for (int j=0;j<=min(i,k);j++) {
                if (j<1) dp[i][j]=(LL)dp[i-1][j]*(son[i]-1)%P*inv[son[i]]%P;
                else dp[i][j]=((LL)dp[i-1][j]*(son[i]-1)%P*inv[son[i]]%P+(LL)dp[i-1][j-1]*1*inv[son[i]]%P)%P;
            }
    
        int ans=dp[n][k];
        for (int i=1;i<=n;i++) ans=((LL)ans*i)%P;
        printf("Case #%d: %d\n",++cas,ans);
    }
    return 0;
}