ACM: 动态规划题 uva 11825

Problem H 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 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<=irepresents the description of node i. The description for node i starts with an integer m (Number of neighbors for node i), followed by mintegers 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: 现在有n个服务, 都运行在n台计算机上, 选择某台计算机上的一项服务停止, 与它相邻的计算机

          上, 该服务也会停止的, 现在要你让尽量多的服务完全停止. (n台计算机上都不运行该服务).

解题思路:

         1. 这题思路, 因为n<=16, 可以采用枚举组合的情况.

             设状态dp[S]表示当选择部分计算机的集合S, 最多可以停止的服务个数.

              状态方程: dp[S] = max(dp[S], dp[S0]+1)

              其中S0是S的子集, 并且满足S0集合可以覆盖全部的计算机.

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 20

int n;
int p[MAX], cover[1<<MAX];
int dp[1<<MAX];

inline int max(int a, int b)
{
    return a > b ? a : b;
}

int main()
{
//    freopen("input.txt", "r", stdin);
    int caseNum = 1;
    while(scanf("%d", &n) != EOF)
    {
        if(n == 0) break;
       
        int num, v;
        for(int i = 0; i < n; ++i)
        {
            scanf("%d", &num);
            p[i] = (1<<i);
            for(int j = 0; j < num; ++j)
            {
                scanf("%d", &v);
                p[i] |= (1<<v);
            }
        }
       
        for(int i = 0; i < (1<<n); ++i)
        {
            cover[i] = 0;
            for(int j = 0; j < n; ++j)
            {
                if(i&(1<<j))
                    cover[i] |= p[j];
            }
        }
       
        dp[0] = 0;
        int all = (1<<n)-1;
        for(int s = 1; s < (1<<n); ++s)
        {
            dp[s] = 0;
            for(int s0 = s; s0 != 0; s0 = (s0-1)&s)
            {
                if(cover[s0] == all)
                    dp[s] = max(dp[s], dp[s^s0]+1);
            }
        }
        printf("Case %d: %d\n", caseNum++, dp[all]);
    }
    return 0;
}