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 $\leq H \leq 50000)H(1≤H≤50000)镑干草.$

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

帮助约翰找到最小的开销来满足需要，即采购到至少

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

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;
}

USACO Buying Hay