UVA 10817——Headmaster's Headache(DP)

最新推荐文章于 2018-03-14 22:31:08 发布

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本文探讨了Springfield学校校长在招聘教师时面临的挑战，如何在满足每个科目至少两位教师授课的同时，实现成本最小化。通过分析输入案例，提出了一种策略来解决这一问题，并给出了具体的最小成本解决方案。

Problem D: Headmaster's Headache
Time limit: 2 seconds

The headmaster of Spring Field School is considering employing some new teachers for certain subjects. There are a number of teachers applying for the posts. Each teacher is able to teach one or more subjects. The headmaster wants to select applicants so that each subject is taught by at least two teachers, and the overall cost is minimized.

Input

The input consists of several test cases. The format of each of them is explained below:

The first line contains three positive integers S, M and N. S (≤ 8) is the number of subjects, M (≤ 20) is the number of serving teachers, and N (≤ 100) is the number of applicants.

Each of the following M lines describes a serving teacher. It first gives the cost of employing him/her (10000 ≤ C ≤ 50000), followed by a list of subjects that he/she can teach. The subjects are numbered from 1 to S. You must keep on employing all of them. After that there are N lines, giving the details of the applicants in the same format.

Input is terminated by a null case where S = 0. This case should not be processed.

Output

For each test case, give the minimum cost to employ the teachers under the constraints.

Sample Input

Sample Output

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <queue>
#include <map>
using namespace std;

const int N = 100 + 10;
const int MOD = 1e9 + 7;
const int INF = 1e9;

int n,m,s;
int dp[N][(1<<16)+2];
int T[N];
int C[N];
int state,cost;

void init()
{
    state=0;
    cost=0;
    memset(T,0,sizeof(T));
    memset(C,0,sizeof(C));

    char c;
    int t;
    for(int i=0;i<m;i++)
    {
        scanf("%d",&t);
        cost += t;
        while((c = getchar()) != '\n')
        {
            if(c==' ') continue;
            else t=c-'0';
            if(state & 1<<(t-1)) state |= 1<<(t-1+s);
            else state |= 1<<(t-1);
        }
    }
    for(int i=0;i<n;i++)
    {
        scanf("%d",&C[i]);
        while((c = getchar()) != '\n')
        {
            if(c==' ') continue;
            else t=c-'0';

            T[i] |= 1<<(t-1);
        }
    }
}

void solve()
{
    memset(dp,-1,sizeof(dp));
    int limit = (1<<(2*s)) - 1;

    dp[0][state] = cost;

    for(int i=0;i<n;i++)
    {
        for(int j=state;j<=limit;j++)
        {
            if(dp[i][j]==-1) continue;

            int k = j&T[i];
            k = k<<s;
            int st = j|T[i];
            st|=k;
            st %= 1<<(s*2);

            if(dp[i+1][st]==-1)
                dp[i+1][st] = dp[i][j]+C[i];
            else
                dp[i+1][st] = min(dp[i][j]+C[i],dp[i+1][st]);

            if(dp[i+1][j]==-1)
                dp[i+1][j]=dp[i][j];
            else
                dp[i+1][j]=min(dp[i+1][j],dp[i][j]);
        }
    }
    printf("%d\n",dp[n][limit]);
}

int main()
{
#ifdef tyh
    freopen("in.txt","r",stdin);
#endif // tyh
    while(scanf("%d%d%d",&s,&m,&n)!=EOF && n+m+s)
    {
        init();
        solve();
    }
    return 0;
}