HDU

题意：n个工作，m个人完成，每个工作有ci个阶段，一个人只能选择一种工作完成，可以不选，且只能完成该工作中与自身标号相同的工作阶段，问最多能完成几种工作。

分析：

1、如果一个工作中的某个工作阶段没有任何一个人能完成，则这个工作是不可完成的。

2、预处理每个人能完成各工作的各工作阶段。

3、剪枝：如果当前完成工作数+剩余人数<=ans则剪枝。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 10 + 10;
const int MAXT = 10000 + 10;
using namespace std;
vector<int> t[MAXN], project[MAXN];
vector<int> engineer[MAXN];
int status[MAXN];
int work[MAXN][MAXN];
int ans;
int n, m;
map<int, int> mp;
int projectnum;
void dfs(int cur, int done){
    if(done + m - (cur - 1) <= ans) return;
    if(cur == m + 1){
        ans = max(ans, done);
        return;
    }
    dfs(cur + 1, done);
    for(int i = 1; i <= projectnum; ++i){
        int len = project[i].size();
        if(status[i] == (1 << len) - 1) continue;
        int tmp = status[i];
        status[i] |= work[cur][i];
        if(status[i] == (1 << len) - 1){
            dfs(cur + 1, done + 1);
        }
        else{
            dfs(cur + 1, done);
        }
        status[i] = tmp;
    }
}
void have(int x, int y){
    int len1 = engineer[x].size();
    int len2 = project[y].size();
    for(int i = 0; i < len2; ++i){
        bool ok = false;
        for(int j = 0; j < len1; ++j){
            if(engineer[x][j] == project[y][i]){
                ok = true;
                break;
            }
        }
        if(ok) work[x][y] |= (1 << i);
    }
}
int main(){
    int T;
    scanf("%d", &T);
    int kase = 0;
    while(T--){
        mp.clear();
        ans = 0;
        for(int i = 0; i < MAXN; ++i){
            project[i].clear();
            engineer[i].clear();
            t[i].clear();
        }
        memset(work, 0, sizeof work);
        memset(status, 0, sizeof status);
        scanf("%d%d", &n, &m);
        int k, x;
        for(int i = 1; i <= n; ++i){
            scanf("%d", &k);
            for(int j = 0; j < k; ++j){
                scanf("%d", &x);
                t[i].push_back(x);
            }
        }
        for(int i = 1; i <= m; ++i){
            scanf("%d", &k);
            for(int j = 0; j < k; ++j){
                scanf("%d", &x);
                mp[x] = 1;
                engineer[i].push_back(x);
            }
        }
        projectnum = 0;
        for(int i = 1; i <= n; ++i){
            int l = t[i].size();
            bool ok = true;
            for(int j = 0; j < l; ++j){
                if(!mp.count(t[i][j])){
                    ok = false;
                    break;
                }
            }
            if(ok) project[++projectnum] = t[i];
        }
        for(int i = 1; i <= m; ++i){
            for(int j = 1; j <= projectnum; ++j){
                have(i, j);
            }
        }
        dfs(1, 0);
        printf("Case #%d: %d
", ++kase, ans);
    }
    return 0;
}