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

题目

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.

题解

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]);

代码

/*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;

}

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