UVA 11825 - Hackers' Crackdown(dp+状态压缩) - c++编程基础

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 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

Output for Sample Input

3

2 1 2

2 0 2

2 0 1

4

1 1

1 0

1 3

1 2

0

Case 1: 3

Case 2: 2

题意：N台电脑，现在有N种服务，现在你可以在每台电脑终止一项服务，他和他相邻的电脑都会被关闭，如果一项服务在所有电脑都没运行，该项服务成功被破坏，问最多能破坏几种服务。

思路：先把每个点集能覆盖到的电脑cover预处理出来。然后枚举每个状态，枚举每个状态的子集，如果该子集能覆盖到全部，状态转移就+1。

状态转移方程 dp[state] = dp[state - substate] + (substate == ((1<
代码：

#include #include #define max(a,b) ((a)>(b) (a):(b)) const int N = 17; const int MAXN = (1<<17); int n, point[N], cover[MAXN], dp[MAXN]; void init() { int num, to, i; memset(dp, 0, sizeof(dp)); memset(point, 0, sizeof(point)); memset(cover, 0, sizeof(cover)); for (i = 0; i < n; i++) { scanf("%d", &num); point[i] = (point[i]|(1< 0; j = (j - 1)&i) { if (cover[j] == (1<