FatMouse' Trade

最新推荐文章于 2022-05-09 20:02:28 发布

runnerzyc

最新推荐文章于 2022-05-09 20:02:28 发布

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分类专栏：贪心问题

本文链接：https://blog.youkuaiyun.com/fastkeeper/article/details/39214079

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本文介绍了一种通过贪心算法解决猫鼠交易问题的方法。该问题描述了如何使用有限数量的猫粮换取最大数量的JavaBeans。通过定义结构体来表示每间仓库，并使用比较函数对这些仓库进行排序，从而实现最优交换。

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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

Sample Output

  
   13.333
31.500

Author

CHEN, Yue

Source

ZJCPC2004

分析：用有的猫粮换取最多的食物，很明显是贪心问题，用结构体做；

代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
struct mouse{
 int pay;
 int get;	
}a[1100];
int cmp(mouse a,mouse b)
{
	return a.get*b.pay>b.get*a.pay;
}
int main()
{
	int n,m,i;
	double num,sum;
	while(~scanf("%d%d",&n,&m))
	{
		if(n==-1&&m==-1)
		break;
		sum=0;num=0;
	 for(i=0;i<m;i++)
	 scanf("%d%d",&a[i].get,&a[i].pay);
	 sort(a,a+m,cmp);
	 for(i=0;i<m;i++)
	 {
	 sum+=a[i].get;
	 num+=a[i].pay;
	 if(num>n)
	{
	 sum-=(num-n)*a[i].get*1.0/a[i].pay;
	 break;
	}
     }
     printf("%.3lf\n",sum*1.0);
	}
	return 0;
}