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; }