UVA 11825 - Hackers' Crackdown(dp+状态压缩)

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 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.

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<<n) - 1));

代码：

#include <stdio.h>
#include <string.h>
#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<<i));
		while (num--) {
			scanf("%d", &to);
			point[i] = (point[i]|(1<<to));
		}
	}
	for (i = 0; i < (1<<n); i++) {
		for (int j = 0; j < n; j++) {
			if (i&(1<<j)) {
				cover[i] = (cover[i]|point[j]);
			}
		}
	}
}

int solve() {
	for (int i = 0; i < (1<<n); i++) {
		for (int j = i; j > 0; j = (j - 1)&i) {
			if (cover[j] == (1<<n) - 1) {
				dp[i] = max(dp[i], dp[i^j] + 1);
			}
		}
	}
	return dp[(1<<n) - 1];
}

int main() {
	int cas = 0;
	while (~scanf("%d", &n) && n) {
		init();
		printf("Case %d: %d\n", ++cas, solve());
	}
	return 0;
}

UVA 11825 - Hackers' Crackdown(dp+状态压缩)

原文：http://blog.csdn.net/accelerator_/article/details/19213533