HDOJ贪心算法

最新推荐文章于 2019-10-31 22:46:50 发布

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最新推荐文章于 2019-10-31 22:46:50 发布

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本文链接：https://blog.youkuaiyun.com/a716121/article/details/46334245

本文探讨了一种独特的交易策略，通过用猫粮交换JavaBean，旨在最大化获得的JavaBean数量。每间房间都包含不同数量的JavaBean和所需猫粮量，通过合理的交易方式，我们可以计算出最大可能获得的JavaBean总量。

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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
5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

Sample Output
13.333
31.500

Author
#include <iostream>
#include<algorithm>
#include<iomanip>
#include<cstdio>
#include<cmath>
using namespace std;
struct Food
{
    double  x,y;
    double rate;

} food[1001];
bool c(Food a,Food b)
{
    return a.rate>b.rate;
}
int main()
{
    int  n,m,i;
    double sum;
  while(scanf("%d%d",&m,&n)&&(m!=-1||n!=-1))
    {
        for(i=0; i<n; i++)
        {
           scanf("%lf%lf",&food[i].x,&food[i].y);
            food[i].rate=food[i].x/food[i].y;
        }
        sort(food,food+n,c);
        sum=0;
        for(i=0; i<n; i++)
        {
            if(m>=food[i].y)
            {
                sum+=food[i].x;
                m-=food[i].y;
            }
            else
            {
                sum+=m*food[i].x/food[i].y;
                break;
            }
        }
       printf("%.3f\n",sum);
    }
    return 0;
}