BZOJ 1618: [Usaco2008 Nov]Buying Hay 购买干草

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题目

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1618: [Usaco2008 Nov]Buying Hay 购买干草

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 679  Solved: 347
[Submit][Status]

Description

    约翰的干草库存已经告罄，他打算为奶牛们采购日(1≤日≤50000)磅干草．

    他知道N(1≤N≤100)个干草公司，现在用1到N给它们编号．第i个公司卖的干草包重量为Pi(1≤Pi≤5000)磅，需要的开销为Ci(l≤Ci≤5000)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H磅干草．

Input

    第1行输入N和日，之后N行每行输入一个Pi和Ci．

Output

 

    最小的开销．

Sample Input

2 15
3 2
5 3

Sample Output

9


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

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题解

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f[i]表示购买至少i的干草的花费，f[remin(v+p[i],h)]=min(f[remin(v+p[i],h)],f[v]+c[i]);

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代码

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/*Author:WNJXYK*/
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
using namespace std;

#define LL long long
#define Inf 2147483647
#define InfL 10000000000LL

inline void swap(int &x,int &y){int tmp=x;x=y;y=tmp;}
inline void swap(LL &x,LL &y){LL tmp=x;x=y;y=tmp;}
inline int remin(int a,int b){if (a<b) return a;return b;}
inline int remax(int a,int b){if (a>b) return a;return b;}
inline LL remin(LL a,LL b){if (a<b) return a;return b;}
inline LL remax(LL a,LL b){if (a>b) return a;return b;}

const int Maxn=50000;
const int N=100;

int f[Maxn+10];
int p[N+10];
int c[N+10];
int n,h;
int main(){
	scanf("%d%d",&n,&h);
	for (int i=1;i<=n;i++) scanf("%d%d",&p[i],&c[i]);
	memset(f,127,sizeof(f));
	f[0]=0;
	for (int i=1;i<=n;i++){
		for (int v=0;v<=h;v++){
			f[remin(v+p[i],h)]=remin(f[remin(v+p[i],h)],f[v]+c[i]);
		}
	}
	printf("%d\n",f[h]);
	return 0;
}