Hackers' Crackdown UVA - 11825

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

题解：

　　这个题目首先把模型抽象出来，n个集合，将每个集合分组，使得每组集合元素的并的于全集，求最多可以分成多少组。

　　首先集合有关，我们数据范围16，我们首先想到子集dp，设dp[S]，表示集合S中各个元素的并等于全集的数量，那么转移十分好转dp[S]=dp[S-S次]+1，S次 是S的子集，且S次的并等于全集。

　　但实现起来还是十分麻烦，首先我们可以用二斤制压缩来表示一个电脑集合p[i]，p[now]|=1<<x;其次就是集合取并集，我们用“|”来实现，就是如果包括这个元素i那么就|=p[i]，最后是枚举子集，枚举集合S的子集S0我们可以for(int s0=s;s0;s0=(s0-1)&s),这个还比较玄学吧。

代码：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<stdio.h>
#include<stdlib.h>
#define MAXN 20
using namespace std;
int n,m;
int dp[<<MAXN],cover[<<MAXN],p[MAXN];
 
void cl(){
    memset(p,,sizeof(p));
    memset(dp,,sizeof(dp));
    memset(cover,,sizeof(cover));
}
 
int main(){
    int ca=;
    while(){
        cl();
        scanf("%d",&n);
        if(!n) break;
        for(int now=;now<n;now++){
            scanf("%d",&m);
            for(int i=;i<=m;i++){
                int x;scanf("%d",&x);
                p[now]|=<<x;
            }
            p[now]|=<<now;
        }
        int all=(<<n)-;
        for(int s=;s<=all;s++){
            for(int i=;i<n;i++)
            if(s&(<<i)) cover[s]|=p[i];
        }
        for(int s=;s<=all;s++){
            for(int s0=s;s0;s0=(s0-)&s){
                if(cover[s0]==all) dp[s]=max(dp[s],dp[s^s0]+);
            }
        }
        printf("Case %d: %d\n",++ca,dp[all]);
    }
}