Hackers’ Crackdown - UVa11825 状压dp-优快云博客

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/41012517

本文探讨了一个黑客如何通过有限的攻击次数最大化瘫痪系统服务的数量。在一个由N个节点组成的分布式系统中，每个节点运行相同的N项服务。黑客的目标是利用特定漏洞攻击尽可能多的服务。文章介绍了解决方案的思路及AC代码。

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Problem H

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 bym 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

2 1 2

2 0 2

2 0 1

1 1

1 0

1 3

1 2

Case 1: 3

Case 2: 2

题意：有n个电脑，一起执行1-n种工程，对于每台电脑，你有一次机会使其不能工作某一项工程，同时与它直接连接的电脑有收到该影响。如果一项工程没有电脑执行，那么这项工程就瘫痪了，问最多能瘫痪多少工程。

思路：首先每天电脑及其直接连接的先状压一下，表示使哪些瘫痪。S表示当前已经使哪些电脑瘫痪过的最大损坏工程数，如果S-S2可以瘫痪一个工程的话，那么dp[S]=max(dp[S],dp[S2]+1)。这里还有一个求子集的实现技巧见代码。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m,pow2[20],val[20],c[100010],dp[100010];
int main()
{
    int t=0,i,j,k,total;
    pow2[0]=1;
    for(i=1;i<=16;i++)
       pow2[i]=pow2[i-1]*2;
    while(~scanf("%d",&n) && n>0)
    {
        for(i=0;i<n;i++)
        {
            val[i]=pow2[i];
            scanf("%d",&m);
            for(j=1;j<=m;j++)
            {
                scanf("%d",&k);
                val[i]+=pow2[k];
            }
        }
        total=pow2[n]-1;
        for(i=1;i<=total;i++)
        {
            c[i]=0;
            for(j=0;j<n;j++)
               if(i&pow2[j])
                 c[i]|=val[j];
        }
        for(i=1;i<=total;i++)
        {
            dp[i]=0;
            for(k=i;k>0;k=(k-1)&i)
               if(c[k]==total)
                 dp[i]=max(dp[i],dp[i^k]+1);
        }
        printf("Case %d: %d\n",++t,dp[total]);
    }

}