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
题目大意：（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）
输入格式：输入包含多组数据。每组数据的第一行为整数n（1<=n<=16）；以下n行每行描述一台计算机的相邻计算机，其中第一个数m为相邻计算机个数，接下来的m个整数位这些计算机的编号。输入结束标志为n=0。
输出格式：对于每组数据，输出完全瘫痪的服务器的最大数量。
分析：
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，即在代码中，每个集合Pi 实际上是一个非负整数。输入的部分代码如下：
for(int i=;i<n;i++){
int m,x;
scanf("%d",&m);
P[i] = <<i;
while(m--) { scanf("%d",&x); P[i] |= (<<x); }
}
为了方便，我们用cover(S)表示若干Pi 的集合S中所有Pi 的并集（二级制表示），即这些Pi 在数值上“按位或”。
for(int S = ; S < (<<n); S++){
cover[S] = ;
for(int i=;i<n;i++)
if(S & (<<i)) cover[S] |= P[i];
}
想到这样的动态规划：用f(S)表示子集S最多可以分成多少组，则
f(S) = max{f(S0)|S0是S的子集，cover[S0]等于全集}+1
这里有一个重要的技巧：枚举S的子集S0。详见下面代码。
int ALL = (<<n) - ;
for(int S = ;S< (<<n); S++){
f[S] = ;
for(int S0 = S; S0; S0 = (S0-)&S)
if(cover[S0] == ALL) f[S] = max(f[S], f[S^S0]+);
}
printf("Case %d: %d\n",++kase,f[ALL]);
完整代码如下：
#include<cstdio>
#include<algorithm>
using namespace std; const int maxn = ;
int n, P[maxn], cover[<<maxn], f[<<maxn];
int main() {
int kase = ;
while(scanf("%d", &n) == && n) {
for(int i = ; i < n; i++) {
int m, x;
scanf("%d", &m);
P[i] = <<i;
while(m--) { scanf("%d", &x); P[i] |= (<<x); }
}
for(int S = ; S < (<<n); S++) {
cover[S] = ;
for(int i = ; i < n; i++)
if(S & (<<i)) cover[S] |= P[i];
}
f[] = ;
int ALL = (<<n) - ;
for(int S = ; S < (<<n); S++) {
f[S] = ;
for(int S0 = S; S0; S0 = (S0-)&S)
if(cover[S0] == ALL) f[S] = max(f[S], f[S^S0]+);
}
printf("Case %d: %d\n", ++kase, f[ALL]);
}
return ;
}
注意：位运算符的优先级比较低，注意加括号