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  • toj 3098 Event Planning

    3098.   Event Planning
    Time Limit: 5.0 Seconds   Memory Limit: 65536K
    Total Runs: 134   Accepted Runs: 67    Multiple test files



    As you didn't show up to the yearly general meeting of the Nordic Club of Pin Collectors, you were unanimously elected to organize this years excursion to Pin City. You are free to choose from a number of weekends this autumn, and have to find a suitable hotel to stay at, preferably as cheap as possible.

    You have some constraints: The total cost of the trip must be within budget, of course. All participants must stay at the same hotel, to avoid last years catastrophe, where some members got lost in the city, never being seen again.

    Input

    The first line of input consists of four integers: 1 ≤ N ≤ 200, the number of participants, 1 ≤ B ≤ 500000, the budget, 1 ≤ H ≤ 18, the number of hotels to consider, and 1 ≤ W ≤ 13, the number of weeks you can choose between. Then follow two lines for each of the H hotels. The first gives 1 ≤ p ≤ 10000, the price for one person staying the weekend at the hotel. The second contains W integers, 0 ≤ a ≤ 1000, giving the number of available beds for each weekend at the hotel.

    Output

    Output the minimum cost of the stay for your group, or "stay home" if nothing can be found within the budget.

    Sample Input

    3 1000 2 3
    200
    0 2 2
    300
    27 3 20

    Sample Output

    900

    Sample Input 2

    5 2000 2 4
    300
    4 3 0 4
    450
    7 8 0 13
    

    Sample Output 2

    stay home
    


    Source: Nordic Collegiate Contest 2008
    Submit   List    Runs   Forum   Statistics

    #include <iostream>
    #include 
    <algorithm>
    #define MAX 10000
    using namespace std;
    int data[MAX];
    bool comp(int a,int b)
    {
        
    return a<b;
    }
    int n,b,h,w;
    int main()
    {
        
    int i,j,a;
        
    while(scanf("%d%d%d%d",&n,&b,&h,&w)!=EOF)
        {
            
    int k=0,m;
            
    for(i=0;i<h;i++)
            {
                scanf(
    "%d",&a);
                
    for(j=0;j<w;j++)
                {
                    scanf(
    "%d",&m);
                    
    if(m<n)
                        
    continue;
                    
    if(n*a>b)
                        
    continue;
                    data[k
    ++]=n*a;
                }
            }
            sort(data,data
    +k,comp);
            
    if(k==0)
                printf(
    "stay home\n");
            
    else
                printf(
    "%d\n",data[0]);
        }
        
    return 0;
    }
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  • 原文地址:https://www.cnblogs.com/forever4444/p/1471066.html
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