HDU1009——FatMouse' Trade

最新推荐文章于 2024-08-22 20:10:34 发布

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本文介绍了一个基于贪心算法的问题解决方法，通过结构体排序实现最优交换策略，帮助FatMouse利用有限的猫粮换取尽可能多的JavaBeans。

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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

分析：
贪心算法——找最优解，我的思路是，运用结构体，运用sort函数将数据从大到小排序，先换: 7 2; 再换 5 2；最后用剩下的1个，换1.33的javabeen。
，然后问题就迎刃而解了。。

#include<algorithm>
#include<iostream>
#include<stdio.h>
using namespace std;//
struct node
{
    int x;
    int y;
    double z;
}p[1005];//this is结构体
int cmp(node a,node b)
{
    return a.z>b.z;
}
//不加这部分默认是从小到大排列
int main()
{
    int m,n,i;
    double s;
    while(scanf("%d%d",&m,&n)&&(m!=-1||n!=-1))
    {
        s=0.0;
        for(i=0; i<n; i++)
        {
            scanf("%d%d",&p[i].x,&p[i].y);
            p[i].z=p[i].x*1.0/p[i].y;
        }
        sort(p,p+n,cmp);//————排序；
        for(i=0; i<n; i++)
        {
            if(m>=p[i].y)
            {
                s=s+p[i].x;
                m=m-p[i].y;
            }
            else
            {
                s=s+m*p[i].z;
                break;
            }
        }

        printf("%.3f\n",s);
    }
    return 0;
}
//OK