HDU-1009 FatMouse' Trade

http://acm.hdu.edu.cn/showproblem.php?pid=1009

dp算法，最主要是 结构体，比例从大到小。

贪心算法HDU1009 FatMouse' Trade

题目大意：

老鼠有M磅猫食。有N个房间，每个房间前有一只猫，房间里有老鼠最喜欢的食品JavaBean,J[i]。若要引开猫，必须付出相应的猫食F[i]。当然这只老鼠没必要每次都付出所有的F[i]。若它付出F[i]的a%，则得到J[i]的a%。求老鼠能吃到的做多的JavaBean。

解题思路：

老鼠要获得最多的食品，就要用最小的猫食换取最多的猫食，这就要求J[i]/F[i]的比例要大。J[i]/F[i]的比例越大，证明在这个房间，小鼠付出得到的收获最有价值。于是我们将设置结构体，结构体里设置percent放置J[i]/F[i]。然后对结构体数组进行排序。依次按比例排序的付出猫食，即可。

例子：

输入

5 3

7 2

4 3

5 2

按J[i]/F[i]排序后：

7 2

5 2

4 3

在第一排数据上，小鼠付出2个猫食换得7个JavaBean。

在第二排数据上，小鼠付出2个猫食换得5个JavaBean。

在第三排数据上，小鼠只剩下1个猫食，便用这一个猫食换取1/3*4个JavaBean。

所以，总共换得13.333个JavaBan。

                       FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 37762    Accepted Submission(s): 12490

Problem Description

FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean. The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

 

Input

The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1's. All integers are not greater than 1000.

 

Output

For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.

 

Sample Input

5 3

7 2

4 3

5 2

20 3

25 18

24 15

15 10

-1 -1

 

 

Sample Output

13.333

31.500

#include<iostream>
#include<algorithm>
using namespace std;
struct Trade
{
    int j,f;
    double percent;
}mouse[3001];
bool cmp(Trade a,Trade b)
{
    return a.percent>b.percent;
}
int main()
{
   int m,n,i,sum,ans,temp;
   double cnt;
   while(~scanf("%d%d",&m,&n))
   {
	     sum=0;ans=0;temp=0;
	   if(m==-1&&n==-1)
		   break;
	   for(i=0;i<n;i++)
	   {
		   scanf("%d%d",&mouse[i].j,&mouse[i].f);
		   mouse[i].percent=(double)mouse[i].j/mouse[i].f;
       }
	   sort(mouse,mouse+n,cmp);
	   for(i=0;i<n;i++)
	   {
		   sum+=mouse[i].f;
		   if(sum<=m)
		   {
			   ans+=mouse[i].j;
			   temp+=mouse[i].f;
			}
		   else
		   {
			   sum=m-temp;
			   break;
		   }
	   }
	   cnt=(double)mouse[i].j*((double)sum/mouse[i].f);
	     printf("%.3lf
",cnt+ans);
   }
   return  0;
}