HDU_1009 FatMouse' Trade 【贪心】

题目信息：

FatMouse' Trade

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 37 24 35 220 325 1824 1515 10-1 -1

 

Sample Output

13.33331.500

解题思路：

基础的贪心算法，按照J[i]/F[i]从大到小排序，i为0~N-1，判断J[i]与M的大小，如果F[i]小于M，SUM就加上J[i]，M刷新为M-J[i]，否则SUM加上M*J[i]/F[i]，M为0。最后将SUM保留三位小数点输出。

代码如下：

#include <iostream>#include <cstdio>#include <cstring>#include <set>#include <algorithm>#include <map>using namespace std;struct food{    double j,f,p;}F[1005];bool cmp(food a,food b){    return a.p>b.p;}int main(){    double m,n;    while(~scanf("%lf%lf",&m,&n))    {        double sum=0;        if(m==-1&&n==-1)            break;        for(int i=0;i<n;i++)        {            scanf("%lf%lf",&F[i].j,&F[i].f);            F[i].p=F[i].j/F[i].f;        }        sort(F,F+(int)n,cmp);        for(int i=0;i<n;i++)        {            if(m>=F[i].f)            {                m-=F[i].f;                sum+=F[i].j;            }            else            {                sum+=F[i].p*m;                break;            }        }        printf("%.3lf\n",sum);    }    return 0;}

注意事项：

• 如果输入用cin会超时（Time Limit Exceeded），改为scanf就可以了
• 多组实例测试，以-1 -1结束