UVA 11825 Hackers’ Crackdown 状压DP枚举子集势

Hackers’ Crackdown

 

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 Nservices. 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

 

题意：

           （黑客的攻击）假设你是一个黑客，侵入了了一个有着n台计算机（编号为0，1，…，n-1）的网络。一共有n种服务，每台计算机都运行着所有的服务。对于每台计算机，你都可以选择一项服务，终止这台计算机和所有与它相邻计算机的该项服务（如果其中一些服务已经停止，则这些服务继续处于停止状态）。你的目标是让尽量多的服务器完全瘫痪（即：没有任何计算机运行该项服务）

题解：

　　　　先把每个点集能覆盖到的电脑cover预处理出来。然后枚举每个状态，枚举每个状态的子集，如果该子集能覆盖到全部，状态转移就+1。

           状态转移方程 dp[state] = dp[state - substate] + (substate == ((1<<n) - 1));

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int N = 17, M = 30005, mod = 1e9+7, inf = 0x3f3f3f3f;
typedef long long ll;
//不同为1，相同为0

int dp[1<<N],c[1<<N],x,n,cas = 0, point[1<<N];
int main() {
    while(scanf("%d",&n) == 1 && n) {
        for(int i=0;i<n;i++) {
            scanf("%d",&x);int j;
            point[i] = (1<<i);
            while(x--) scanf("%d",&j), point[i]|=(1<<j);
        }
        for(int i=0;i<(1<<n);i++) {
            c[i] = 0;
            for(int j=0;j<n;j++) {
                if(i&(1<<j)) c[i]|=point[j];
            }
        }
        for(int i=1;i<(1<<n);i++) {
            dp[i] = 0;
            for(int j=i;j;j = (j-1)&i) {
                if(c[j] == (1<<n) - 1) dp[i] = max(dp[i], dp[i^j] + 1);
            }
        }
        printf("Case %d: %d\n", ++cas, dp[(1<<n) - 1]);
    }
}