[BZOJ1618][Usaco2008 Nov]Buying Hay 购买干草

1618: [Usaco2008 Nov]Buying Hay 购买干草

Time Limit: 5 Sec  Memory Limit: 64 MB Submit: 1224  Solved: 639 [Submit][Status][Discuss]

Description

约翰的干草库存已经告罄,他打算为奶牛们采购H(1≤H≤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H，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.

 

完全背包

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
char buf[10000000], *ptr = buf - 1;
inline int readint(){
    int n = 0;
    char ch = *++ptr;
    while(ch < '0' || ch > '9') ch = *++ptr;
    while(ch <= '9' && ch >= '0'){
        n = (n << 1) + (n << 3) + ch - '0';
        ch = *++ptr;
    }
    return n;
}
int dp[50000 + 10];
int main(){
    fread(buf, sizeof(char), sizeof(buf), stdin);
    memset(dp, 0x3f, sizeof dp);
    dp[0] = 0;
    int n, H;
    n = readint();
    H = readint();
    for(int p, c, i = 1; i <= n; i++){
        p = readint();
        c = readint();
        for(int i = 1; i <= p && i <= H; i++)
            dp[i] = min(dp[i], c);
        for(int i = p + 1; i <= H; i++)
            dp[i] = min(dp[i], dp[i - p] + c);
    }
    printf("%d\n", dp[H]);
    return 0;
}