UVA 11825 dp、状态压缩、二进制法表示集合

http://vjudge.net/vjudge/contest/view.action?cid=53516#problem/D

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. 
 
Sample Input 
3 
2 1 2 
2 0 2 
2 0 1 
4 
1 1 
1 0 
1 3 
1 2 
0 

Output for Sample Input 

Case 1: 3 
Case 2: 2 
题目大意：（大白书p69）

           假设你是一个黑客，侵入了一个有着n台计算机（编号为0,1,2......n-1）的网络。一共有n种服务，每台计算机都运行着所有服务。对于每台计算就，你都可以选择一项服务，终止这台计算机的该项服务（如果其中一些服务已经终止，则这些服务继续处于停止的状态）。你的目标是让尽量多的服务完全瘫痪（即没有任何计算机运行该项服务）

解题思路：

          本题的数学模型是：把n个集合p1,p2,......pn分成尽量多组，使得每组中所有集合的并集等于全集。这里的集合pi就是计算机i及其相邻计算机的集合，每组对应于题目中的一项服务，注意到n很小，可以用二进制法表示这些集合，在代码中，每个集合pi实际上是一个非负整数。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=1<<16+2;
int f[maxn],cover[maxn],p[maxn];
int n;
int main()
{
    int tt=0;
    while(~scanf("%d",&n))
    {
        if(n==0)
            break;
        //二进制法表示集合p[i];
        for(int i=0;i<n;i++)
        {
            int x,m;
            scanf("%d",&m);
            p[i]=1<<i;
            while(m--)
            {
                scanf("%d",&x);
                p[i]|=(1<<x);
            }
        }
        //cover(s)表示若干pi的集合s中所有pi的并集（二进制表示），即：这些pi在数值上“按位或”
        for(int s=0;s<(1<<n);s++)
        {
            cover[s]=0;
            for(int i=0;i<n;i++)
            {
                if(s&(1<<i))
                    cover[s]|=p[i];
            }
        }
        //dp：用F[s]表示子集s最多可以分成多少组，则状态转移方程为：f(s)=max(f(s-s0)+1,f(s)).s0是s的子集，cover[s0]等于全集
        f[0]=0;
        int all=(1<<n)-1;
        for(int s=1;s<(1<<n);s++)
        {
            f[s]=0;
            for(int s0=s;s0;s0=(s0-1)&s)//枚举s的子集s0.
            {
                if(cover[s0]==all)
                    f[s]=max(f[s],f[s^s0]+1);//f[s^0]即为：f(s-s0).
            }
        }
        printf("Case %d: %d\n",++tt,f[all]);
    }
    return 0;
}