FatMouse' Trade （贪心）

最新推荐文章于 2019-08-11 08:49:19 发布

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解决一个关于使用不同比例交换资源的问题，通过贪心算法找出最优解，实现利益最大化。

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FatMouse' Trade

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.

InputThe 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.
OutputFor 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

题目大意：

胖老鼠有m磅猫粮和猫换食物，有n个房间，每个房间他可以用bi磅猫粮换ai磅食物，问用完全部猫粮最多可换多少食物

思路：（贪心）
以兑换率进行排序，从最划算的房间开始换，直到全部换完

代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
    int a,b;
    double c;
}j[1005];
bool cmp(node x,node y)
{
    return x.c<y.c;
}
int main()
{
    int m,n;
    while(~scanf("%d%d",&m,&n))
    {
        if(m==-1&&n==-1)
            break;
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&j[i].a,&j[i].b);
            j[i].c=j[i].b*1.0/j[i].a;
        }
        sort(j,j+n,cmp);
        double sum=0;
        for(int i=0;i<n;i++)
        {
            if(m>=j[i].b)
            {
                m-=j[i].b;
                sum+=j[i].a;
            }
            else
            {
                sum+=m*j[i].a*1.0/j[i].b;
                m=0;
            }
            if(m==0)
                break;
        }
        printf("%.3lf\n",sum);
    }
    return 0;
}