Hackers’ Crackdown 
Input: Standard Input

Output: Standard Output

Miracle Corporations has a number of system services running in a distributed computer system which is a prime target for hackers. The system is basically a set of N computer nodes with each of them running a set of N services. Note that, the set of services running on every node is same everywhere in the network. A hacker can destroy a service by running a specialized exploit for that service in all the nodes.

One day, a smart hacker collects necessary exploits for all these N services and launches an attack on the system. He finds a security hole that gives him just enough time to run a single exploit in each computer. These exploits have the characteristic that, its successfully infects the computer where it was originally run and all the neighbor computers of that node.

Given a network description, find the maximum number of services that the hacker can damage.

Input

There will be multiple test cases in the input file. A test case begins with an integer N (1<=N<=16), the number of nodes in the network. The nodes are denoted by 0 to N - 1. Each of the following N lines describes the neighbors of a node. Line i (0<=i<N) represents the description of node i. The description for node i starts with an integer m (Number of neighbors for node i), followed by m integers in the range of 0 to N - 1, each denoting a neighboring node of node i.

The end of input will be denoted by a case with N = 0. This case should not be processed.

Output

For each test case, print a line in the format, “Case X: Y”, where X is the case number & Y is the maximum possible number of services that can be damaged.

Problemsetter: Mohammad Mahmudur Rahman

Special Thanks  Manzurur Rahman Khan

题目大意：

有n台计算机，每台计算机运行n个不同进程，现在你可以每台机器上停止一个服务，而且你停止了1台机器上的这个服务的同时，其相连机器上的这个服务也会停止，再告诉 你每台机器相连的机器，当一个所有机器上的这个服务都停止了，那么这个服务才算没有被运行，问你最多多少个服务没有 被运行？

解题思路：

其实就是把这些机器分成最多的子集合集合，每个子集合合并起来能够影响全部，这样就能解决问题。这样枚举的状态就是2^16次方。

解题代码：

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

const int maxn=(1<<16)+10;
int n,all,dp[maxn],a[maxn],value[maxn];

void initial(){
    all=(1<<n)-1;
    for(int i=0;i<=all;i++) dp[i]=-1;
}

void input(){
    for(int i=0;i<n;i++){
        int m,x;
        scanf("%d",&m);
        a[i]=(1<<i);
        while(m-- >0){
            scanf("%d",&x);
            a[i]|=(1<<x);
        }
    }
    for(int i=0;i<=all;i++){
        value[i]=0;
        for(int t=0;t<n;t++){
            if( i&(1<<t) ){
                value[i]|=a[t];
            }
        }
    }
}

int DP(int sum){
    if(value[sum]!=all ) return 0;
    if(dp[sum]!=-1) return dp[sum];
    int ans=0;
    for(int x=sum;x!=0;x=(x-1)&sum ){
        if(value[x]==all){
            if(1+DP(sum-x)>ans) ans=1+DP(sum-x);
        }
    }
    return dp[sum]=ans;
}

int main(){
    int casen=0;
    while(scanf("%d",&n)!=EOF && n!=0 ){
        initial();
        input();
        printf("Case %d: %d\n",++casen,DP(all) );
    }
    return 0;
}