讲题

#include <stdio.h>
#define maxn 1005
#define mod 1000000007
int cas,t = 1,road[maxn][maxn],n,k,ra,rb;
long long A[maxn],dp[maxn][maxn],size[maxn],ans;       /*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;
    A[0] = 1;
    for(i = 1;i < maxn;i ++)
    {
        A[i] =  A[i-1] * i % mod;
    }
}
void init1()
{
    int i,j;
    for(i = 1;i <= n;i ++)
    {
        size[i] = 1;
        for(j = 1;j <= n;j ++)
            dp[i][j] = road[i][j] = 0;
    }
}
long long fast_pow(long long x, int n) {
    long long ret = 1;
    while (n) {
        if (n&1) ret = ret * x % mod;
        n >>= 1;
        x = x * x % mod;
    }
    return ret;
}

long long inv(long long x) {
    return fast_pow(x, mod-2);
}

/*得到以每个点为根的子树的节点数*/
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()
{
    int i,j;
    long long p,q;
    scanf("%d",&cas);
    init();
    while(cas --)
    {
        scanf("%d %d",&n,&k);
        init1();
        for(i = 1;i < n;i ++)
        {
            scanf("%d %d",&ra,&rb);
            road[ra][rb] = road[rb][ra] = 1;
        }
        LookRoad(1);
        /*逆元代替分数*/
        dp[0][0] = 1;
        dp[0][1] = 0;
        for(i = 1;i <= n;i ++)
        {
            p = inv(size[i]);                                     /*该点是最大点的概率*/
            q = (size[i]-1)*p%mod;                                /*该点不是最大点的概率*/
            dp[i][0] = dp[i-1][0] * q % mod;
            for(j = 1;j <= i;j ++)
            {
                    dp[i][j] = (dp[i-1][j-1]*p%mod + dp[i-1][j]*q%mod)%mod;
            }
        }
        ans = dp[n][k] * A[n] % mod;
        printf("Case #%d: %lld
",t++,ans);
    }
    return 0;
}