J. Leader in Tree Land

Posted 2020-06-21

#include <iostream>

using namespace std;
int cas,t = 1,road[1001][1001],n,k,ra,rb,size[1001];
long long sum,dp[1001][1001];       //dp[i][j]表示从1到i节点有j个节点是其字数的最大值
void init()
{
    int i,j;
    sum = 1;
    for(i = 1;i <= n;i ++)
    {
        sum = (sum*i)%1000000007;
        size[i] = 1;
        for(j = 1;j <= n;j ++)
            dp[i][j] = road[i][j] = 0;
    }
}
void LookRoad(int r)
{
    int i;
    for(i = 1;i <= n;i ++)
    {
        if(road[r][i])
        {
            road[r][i] = road[i][r] = 0;
            LookRoad(i);
            size[r] += size[i];
        }
    }
}
int main()
{
    cin >> cas;
    while(cas --)
    {
        cin >> n >> k;
        init();
        for(int i = 1;i < n;i ++)
        {
            cin >> ra >> rb;
            road[ra][rb] = road[rb][ra] = 1;
        }
        LookRoad(1);
        dp[1][0] = ((n-1)*sum)/n;
        dp[1][1] = sum/n;
        for(int i = 2;i <= n;i ++)
        {
            for(int j = 0;j <= i;j ++)
            {
                if(j > 0)
                    dp[i][j] = dp[i-1][j-1]/size[i] + (dp[i-1][j]*(size[i]-1))/size[i];
                else
                    dp[i][0] = (dp[i-1][0] *(size[i]-1))/size[i];
            }
        }
        cout << "case #" << t ++ << ": " << dp[n][k] << endl;
    }
    return 0;
}