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=0;i
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 = 0; S < (1<
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 = (1<
完整代码如下:
1 #include2 #include 3 using namespace std; 4 5 const int maxn = 16; 6 int n, P[maxn], cover[1<
注意:位运算符的优先级比较低,注意加括号