题目链接：

Fluorescent

Time Limit: 3000/3000 MS (Java/Others)    

Memory Limit: 512000/512000 K (Java/Others)

Problem Description

Matt, a famous adventurer who once defeated a pack of dire wolves alone, found a lost court. Matt finds that there are N fluorescent lights which seem to be the stars from the firmament. What’s more, there are M switches that control these fluorescent lights. Each switch is connected to a group of lights. When Matt touches a switch, all the lights connected to it will change their states (turning the dark on, turning the bright off).

Initially, all the fluorescent lights are dark. For each switch, Matt will touch it with probability 1 .

As a curious gentleman, Matt wants to calculate E[X3], where X represents the number of bright lights at the end, E[X3] represents the expectation of cube of X.

Input

The first line contains only one integer T , which indicates the number of test cases.

For each test case, the first line contains N, M (1 ≤ N, M ≤ 50), denoting the number of fluorescent lights (numbered from 1 to N ) and the number of switches (numbered from 1 to M ).

M lines follow. The i-th line begins with an integer Ki (1 ≤ K_i ≤ N ). K_i distinct integers l_ij(1 ≤ l_ij ≤ N ) follow, denoting the fluorescent lights that the i-th switch controls.

Output

For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y is the answer. To avoid rounding error, the answer you should output is:

E[X³] × 2^M mod (10⁹ + 7)

Sample Input

2

2 2

1 1

2 1 2

3 1

3 1 2 3

Sample Output

Case #1: 10

Case #2: 27

题意：一些灯和一些开关，每个开关都能控制一部分灯，求在开关的所有状态亮灯个数立方和

思路：ans=sigma(x^3) ,把x^3拆开就是（x1+x2+...xn）*（x1+x2+...+xn）*(x1+x2+...+xn),ans=sigma(∑∑∑(xi*xj*xk))=∑∑∑(sigma(使得xixjxk同时亮的状态数))

这样可以避免枚举2^m个状态,因为只有三个灯亮,状态为2^3，所以复杂度约为O(n^3*m))

AC代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int n,m,a[51][51],k[50];
LL dp[51][8];
const LL mod=1e9+7;
LL solve(int u,int v,int w)
{
    memset(dp,0,sizeof(dp));
    dp[0][0]=1;
    for(int i=1;i<=m;i++)
    {
        int tep=0;
        for(int j=1;j<=k[i];j++)
        {
            if(a[i][j]==u)tep+=1;
            if(a[i][j]==v)tep+=2;
            if(a[i][j]==w)tep+=4;
        }
        for(int j=0;j<8;j++)dp[i][j]=dp[i-1][j];
        for(int j=0;j<8;j++)dp[i][j^tep]+=dp[i-1][j],dp[i][j^tep]%=mod;
    }
    return dp[m][7]%mod;
}
int main()
{
    int T,Case=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++)
        {
            scanf("%d",&k[i]);
            for(int j=1;j<=k[i];j++)scanf("%d",&a[i][j]);
        }
        LL ans=0;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                for(int k=1;k<=n;k++)
                    ans=ans+solve(i,j,k),ans%=mod;
        printf("Case #%d: %lld\n",++Case,ans);
    }
    return 0;
}