uva 11825 Hackers' Crackdown（动态规划-状态压缩DP）

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.

                                                 

Sample Input

Output for Sample Input

3

2 1 2

2 0 2

2 0 1

4

1 1

1 0

1 3

1 2

0

Case 1: 3

Case 2: 2

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Problemsetter: Mohammad Mahmudur Rahman

Special Thanks  Manzurur Rahman Khan

题目大意：

有n台计算机，每台计算机运行n个不同进程，现在你可以每台机器上停止一个服务，而且你停止了1台机器上的这个服务的同时，其相连机器上的这个服务也会停止，再告诉 你每台机器相连的机器，当一个所有机器上的这个服务都停止了，那么这个服务才算没有被运行，问你最多多少个服务没有 被运行？

解题思路：

其实就是把这些机器分成最多的子集合集合，每个子集合合并起来能够影响全部，这样就能解决问题。这样枚举的状态就是2^16次方。

解题代码：

#include <iostream>#include <cstdio>using namespace std;const int maxn=(1<<16)+10;int n,all,dp[maxn],a[maxn],value[maxn];void initial(){    all=(1<<n)-1;    for(int i=0;i<=all;i++) dp[i]=-1;}void input(){    for(int i=0;i<n;i++){        int m,x;        scanf("%d",&m);        a[i]=(1<<i);        while(m-- >0){            scanf("%d",&x);            a[i]|=(1<<x);        }    }    for(int i=0;i<=all;i++){        value[i]=0;        for(int t=0;t<n;t++){            if( i&(1<<t) ){                value[i]|=a[t];            }        }    }}int DP(int sum){    if(value[sum]!=all ) return 0;    if(dp[sum]!=-1) return dp[sum];    int ans=0;    for(int x=sum;x!=0;x=(x-1)&sum ){        if(value[x]==all){            if(1+DP(sum-x)>ans) ans=1+DP(sum-x);        }    }    return dp[sum]=ans;}int main(){    int casen=0;    while(scanf("%d",&n)!=EOF && n!=0 ){        initial();        input();        printf("Case %d: %d\n",++casen,DP(all) );    }    return 0;}