UVA11825 Hacker's Crackdown

Hackers' Crackdown

https://vjudge.net/problem/UVA-11825

题目大意：n台电脑，n中服务，每台电脑有全部的服务，每台电脑可以选择一种服务，使得这台电脑和它连接的电脑服务终止。问最多能让多少种服务完全瘫痪，一种服务完全瘫痪是指没有任何一台电脑运行这种服务

每台电脑和它所连的电脑构成一个集合，问题变为有若干集合，将他们进行合并，最多能合并成多少全集。

state[i]表示电脑i构成的集合

uni[S]表示S中电脑的集合的并

dp[S]表示S中电脑的集合最多合并成多少全集

dp[S] = max(dp[S], dp[S - SS]),SS为S子集且uni[SS]为全集

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
inline void swap(int &a, int &b)
{
    int tmp = a;a = b;b = tmp;
}
inline void read(int &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}

const int INF = 0x3f3f3f3f;

int n, uni[1 << 16], state[17], dp[1 << 16], ma, ans, t;

int main()
{
    while(scanf("%d", &n) != EOF && n)
    {
        ++ t;
        memset(state, 0, sizeof(state));
        memset(uni, 0, sizeof(uni));
        memset(dp, 0, sizeof(dp));
        for(register int i = 0;i < n;++ i)
        {
            int tmp1, tmp2;read(tmp1);
            state[i] |= (1 << i); 
            for(register int j = 1;j <= tmp1;++ j)
                read(tmp2), state[i] |= (1 << tmp2);
        }
        ma = 1 << n;ans = 0;
        for(register int S = 0;S < ma;++ S)
            for(register int i = 0;i < n;++ i)
                if(S & (1 << i)) uni[S] |= state[i];
        for(register int S = 0;S < ma;++ S)
            for(register int SS = S;SS;SS = (SS - 1) & S)
                if(uni[SS] == ma - 1) dp[S] = max(dp[S], dp[S ^ SS] + 1);
        printf("Case %d: %d
", t, dp[ma - 1]);
    }
    return 0;
}