UVa 11825 Hackers' Crackdown (状压DP)

题意：给定 n 个计算机的一个关系图，你可以停止每台计算机的一项服务，并且和该计算机相邻的计算机也会终止，问你最多能终止多少服务。

析：这个题意思就是说把 n 台计算机尽可能多的分成一些组，使得每组的的 u 是全集。我们可以用状压DP来解决，先处理输入，然后再处理每个子集，

dp[s] 表示状态为 s 时，最多能终止多少服务，dp[s] = max{ dp[s^s0] +1 }。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = (1<<16) + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int dp[maxn];
int a[20], s[maxn];

int main(){
    int kase = 0;
    while(scanf("%d", &n) == 1 && n){
        for(int i = 0; i < n; ++i){
            scanf("%d", &m);
            a[i] = 1 << i;
            int x;
            while(m--){
                scanf("%d", &x);
                a[i] |= 1 << x;
            }
        }
        for(int i = 0; i < (1<<n); ++i){
            s[i] = 0;
            for(int j = 0; j < n; ++j)
                if(i & (1<<j))  s[i] |= a[j];
        }
        int all = (1<<n) - 1;
        dp[0] = 0;
        for(int i = 1; i < (1<<n); ++i){
            dp[i] = 0;
            for(int j = i; j; j = (j-1)&i)
                if(s[j] == all)  dp[i] = max(dp[i], dp[i^j]+1);
        }
        printf("Case %d: %d\n", ++kase, dp[all]);
    }
    return 0;
}