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 MBSubmit: 641 Solved: 328
[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
3 2
5 3
Sample Output
9
FJ can buy three packages from the second supplier for a total cost of 9.
FJ can buy three packages from the second supplier for a total cost of 9.
HINT
Source
题解:
呵呵。。。又是无限背包。。。费用取负,多算一点即可
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set> 10 #include<queue> 11 #include<string> 12 #define inf 1000000000 13 #define maxn 50000+1000 14 #define maxm 5000+100 15 #define eps 1e-10 16 #define ll long long 17 #define pa pair<int,int> 18 using namespace std; 19 inline int read() 20 { 21 int x=0,f=1;char ch=getchar(); 22 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 23 while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();} 24 return x*f; 25 } 26 int n,m,f[maxn]; 27 int main() 28 { 29 freopen("input.txt","r",stdin); 30 freopen("output.txt","w",stdout); 31 n=read();m=read(); 32 memset(f,128,sizeof(f)); 33 f[0]=0; 34 for(int i=1;i<=n;i++) 35 { 36 int x=read(),y=read();y=-y; 37 for(int j=x;j<=m+maxm;j++)f[j]=max(f[j],f[j-x]+y); 38 } 39 int ans=-inf; 40 for(int i=m;i<=m+maxm;i++)ans=max(ans,f[i]); 41 printf("%d\n",-ans); 42 return 0; 43 }