bzoj1618: [Usaco2008 nov]buying hay buy hay full backpack

title : http://www.lydsy.com/JudgeOnline/problem.php?id=1618
1618: [Usaco2008 nov]buying Hay buys Hay
Time Limit:5 Sec Memory limit:64 MB
submit:939 solved:481
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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.

Ideas :
Bare full backpack;

Code :

#include <iostream>#include <stdio.h>#include <string.h>using namespace STD;intN,m;intdp[55005];intv[5005],w[5005];intMain () {scanf("%d%d", &n,&m); for(intI=1; i<=n;i++) {scanf("%d%d", &w[i],&v[i]); } for(intj=1; j<=m+ the; j + +) dp[j]=99999999; dp[0]=0; for(intI=1; i<=n;i++) for(intj=w[i];j<=m+ the; j + +) {dp[j]=min (dp[j],dp[j-w[i]]+v[i]); }intans=99999999; for(inti=m;i<=m+ the; i++) Ans=min (Ans,dp[i]);printf("%d\n", ans);}

bzoj1618: [Usaco2008 nov]buying hay to buy hay full backpack