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  • USACO Buying Hay

    洛谷 P2918 [USACO08NOV]买干草Buying Hay

    https://www.luogu.org/problem/P2918

    JDOJ 2592: USACO 2008 Nov Silver 1.Buying Hay

    https://neooj.com:8082/oldoj/problem.php?id=2592

    题目描述

    Farmer John is running out of supplies and needs to purchase H (1 <= H <= 50,000) pounds of hay for his cows.

    He knows N (1 <= N <= 100) hay suppliers conveniently numbered 1..N. Supplier i sells packages that contain P_i (1 <= P_i <= 5,000) pounds of hay at a cost of C_i (1 <= C_i <= 5,000) dollars. Each supplier has an unlimited number of packages available, and the packages must be bought whole.

    Help FJ by finding the minimum cost necessary to purchase at least H pounds of hay.

    约翰的干草库存已经告罄,他打算为奶牛们采购H(1 leq H leq 50000)H(1H50000)镑干草.

    他知道N(1 leq Nleq 100)N(1N100)个干草公司,现在用11到NN给它们编号.第ii公司卖的干草包重量 为P_i (1 leq P_i leq 5,000)Pi(1Pi5,000) 磅,需要的开销为C_i (1 leq C_i leq 5,000)Ci(1Ci5,000) 美元.每个干草公司的货源都十分充足, 可以卖出无限多的干草包.

    帮助约翰找到最小的开销来满足需要,即采购到至少HH镑干草.

    输入格式

    * Line 1: Two space-separated integers: N and H

    * Lines 2..N+1: Line i+1 contains two space-separated integers: P_i and C_i

    输出格式

    * Line 1: A single integer representing the minimum cost FJ needs to pay to obtain at least H pounds of hay.

    输入输出样例

    输入 #1
    2 15 
    3 2 
    5 3 
    
    输出 #1
    9 
    

    说明/提示

    FJ can buy three packages from the second supplier for a total cost of 9.

    裸的完全背包。

    出了一些小bug。

    我FSW以后要是再用memset置数组,我当场吃翔。

    这道题一个坑点就是,你可以多买一些草,不一定非得是h,因为公司提供的是干草包,而干草包最多有5000,所以需要把范围多置5000,这是细节。

    然后就是纯套完全背包模板。

    祝大家刷题愉快。

    #include<cstdio>
    #include<algorithm>
    using namespace std;
    int n,h,ans=1e9;//n,种类数,h,总体积
    int m[101],w[101];
    int dp[55001];
    int main()
    {
        for(int i=1;i<=55001;i++)
            dp[i]=1e9;
        scanf("%d%d",&n,&h);
        for(int i=1;i<=n;i++)
            scanf("%d%d",&m[i],&w[i]);
        for(int i=1;i<=n;i++)
            for(int j=m[i];j<=h+5000;j++)
                dp[j]=min(dp[j],dp[j-m[i]]+w[i]);
        for(int i=h;i<=h+5000;i++)
            ans=min(ans,dp[i]);
        printf("%d",ans);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/fusiwei/p/11258326.html
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