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[1<<MAXN],cover[1<<MAXN],p[MAXN]; void cl(){ memset(p,0,sizeof(p)); memset(dp,0,sizeof(dp)); memset(cover,0,sizeof(cover)); } int main(){ int ca=0; while(1){ cl(); scanf("%d",&n); if(!n) break; for(int now=0;now<n;now++){ scanf("%d",&m); for(int i=1;i<=m;i++){ int x;scanf("%d",&x); p[now]|=1<<x; } p[now]|=1<<now; } int all=(1<<n)-1; for(int s=0;s<=all;s++){ for(int i=0;i<n;i++) if(s&(1<<i)) cover[s]|=p[i]; } for(int s=1;s<=all;s++){ for(int s0=s;s0;s0=(s0-1)&s){ if(cover[s0]==all) dp[s]=max(dp[s],dp[s^s0]+1); } } printf("Case %d: %d ",++ca,dp[all]); } }