hdu 1009 FatMouse' Trade(贪心)

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本文探讨了利用贪心算法解决FatMouse与守护仓库的猫交易JavaBean的问题，详细介绍了输入输出规范，算法思路，以及通过排序和循环实现求解过程，最后给出AC代码供读者参考。

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题目来源：hdu 1009 FatMouse’ Trade

FatMouse’ Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 54581 Accepted Submission(s): 18299

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
题目大意：
输入m,n，m可以看做总容量，n组数据，每组数据输入J、F；J可看做该物品的总价值，F可看做该物品所占空间为F。所求的是m空间所能装的最大价值是多少，保留三位小数。
题目分析：
此题意在考查贪心思想，想要获得最大价值，我们就要首先取单位价值最大的，将其取完或者取满空间，若空间较大，则再取剩余的单位价值最大的物品。由此可知，此题需要对物品的单位价值进行排序，即(J/F)值，然后使用贪心即可。
AC代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node{        //用结构体存放该物品的价值、体积、单位价值 
    int j,f;
    double v;
};
node edge[1010];
int cmp(node a,node b)  //对物品的单位价值从大到小进行排序 
{
    return a.v>b.v;
}
int main()
{
    int m,n;
    while(scanf("%d%d",&m,&n)!=EOF&&m!=-1&&n!=-1)
    {
        for(int i=0;i<n;i++)            //输入每件物品的价值及体积，并计算单位体积 
        {
            scanf("%d%d",&edge[i].j,&edge[i].f);
            edge[i].v=(double)edge[i].j/edge[i].f;
        }
        sort(edge,edge+n,cmp);  //进行排序 
        double ans=0;           //用来计算最大总价值 
        for(int i=0;i<n;i++)
        {
            if(m>=edge[i].f)    //若空间较大，则取完此物品，剩余空间为m减去该物品的体积 
            {
                m-=edge[i].f;   
                ans+=edge[i].j; //ans计算当前所取的最大价值 
            }
            else
            {
                ans+=edge[i].v*m;   //空间不足以装完此物品时，就装满剩余空间 
                break;              //此时所装价值为单位价值与所装空间之积，已无法继续装，故跳出 
            }
        }
        printf("%.3lf\n",ans);
    }
    return 0;
}