hdu 1009 FatMouse' Trade(贪心)

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;
}

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hdu 1009 FatMouse' Trade(贪心)