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.

 状压DP、枚举子集

#include <iostream>#include <cstring>  #include <cstdio>using namespace std;  #define N (1<<16)+10int n;int p[20];int cover[N];int dp[N];int main()  {    int i,j,k,iCase=1;    while(scanf("%d",&n),n)    {        for(i=0;i<n;i++)        {            int m,x;            p[i]=1<<i;            scanf("%d",&m);            while(m--)            {                scanf("%d",&x);                p[i]|=(1<<x);            }        }        int MAX=1<<n;        for(j=0;j<MAX;j++)        {            cover[j]=0;            for(i=0;i<n;i++)            {                if(j&(1<<i)) cover[j]|=p[i];            }        }        for(j=0;j<MAX;j++)         //枚举集合j        {            dp[j]=0;            for(k=j;k;k=(k-1)&j)   //枚举子集k            {                if(cover[k]==MAX-1) dp[j]=max(dp[j],dp[j^k]+1);            }        }        printf("Case %d: %d\n",iCase++,dp[MAX-1]);    }    return 0;}

状压DP [Uva 11825] Hackers’ Crackdown