题意:有M个已聘教师,N个候选老师,S个科目,已知每个老师的雇佣费和可教科目,已聘老师必须雇佣,要求每个科目至少两个老师教的情况下,最少的雇佣费用。
分析:
1、为让雇佣费尽可能少,雇佣的老师应教他所能教的所有科目。
2、已聘老师必须选,候选老师可选可不选。
3、dfs(cur, subject1, subject2)---求出在当前已选cur个老师,有一个老师教的科目状态为 subject1,有两个及以上老师教的科目状态为 subject2的情况下,最少的雇佣费用。
dp[cur][subject1][subject2]---在当前已选cur个老师,有一个老师教的科目状态为 subject1,有两个及以上老师教的科目状态为 subject2的情况下,最少的雇佣费用。
4、如果第cur个老师选择,则若他能教的科目和在subject1中所对应的科目都为1,那将该科目变为能被两个老师教。
subject1 & teacher[cur] --- 筛选出这个老师能教的科目中,目前只有一个老师教的科目---让这个老师去教这些科目的话,这些科目就被两个老师教了。
subject2 |= (subject1 & teacher[cur]);---将能被两个老师教的科目并到集合里。
#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 = 120 + 10; const int MAXT = 300 + 10; using namespace std; int cost[MAXN]; int teacher[MAXN]; int dp[MAXN][MAXT][MAXT]; string s; int S, M, N; int dfs(int cur, int subject1, int subject2){ if(cur == M + N){ return subject2 == (1 << S) - 1 ? 0 : INT_INF; } int &ans = dp[cur][subject1][subject2]; if(ans >= 0) return ans; ans = INT_INF; if(cur >= M) ans = dfs(cur + 1, subject1, subject2);//第cur个老师不选 subject2 |= (subject1 & teacher[cur]);//第cur个老师选 subject1 |= teacher[cur];//将这个老师能教的科目中,目前没人教的科目,让他教 ans = min(ans, cost[cur] + dfs(cur + 1, subject1, subject2)); return ans; } int main(){ while(scanf("%d%d%d", &S, &M, &N) == 3){ if(!S && !M && !N) return 0; for(int i = 0; i < M + N; ++i){ scanf("%d", &cost[i]); getline(cin, s); stringstream ss(s); int x; teacher[i] = 0; while(ss >> x){ teacher[i] |= 1 << (x - 1); } } memset(dp, -1, sizeof dp); printf("%d ", dfs(0, 0, 0)); } return 0; }