HDU 1009 FatMouse' Trade(贪心)

题目网址：http://acm.hdu.edu.cn/showproblem.php?pid=1009

题目：

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 79300    Accepted Submission(s): 27358

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

思路：

贪心算法。将数据先处理，算出rate[i]=J[i]/F[i]，根据rate[i]做降序排序，比值大的先喂，得到的JavaBeans总值就最大。

代码：

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
struct node{
    int j,f;
    double rate;
};
vector<node>v;
bool cmp(node x,node y){
    return x.rate>y.rate;
}
int main(){
    int m,n;
    double res;
    while (scanf("%d%d",&m,&n)!=EOF && m!=-1 && n!=-1) {
        res=0;
        v.clear();
        for (int i=0; i<n; i++) {
            node x;
            scanf("%d%d",&x.j,&x.f);
            x.rate=x.j*1.0/x.f;
            v.push_back(x);
        }
        sort(v.begin(), v.end(), cmp);
        for (int i=0; i<n; i++) {
            if(v[i].f<=m){//则喂100%f[i]
                m-=v[i].f;
                res+=v[i].j;
            }else{//将剩余的m全部喂掉
                res+=m*1.0/v[i].f*v[i].j;
                m=0;
                break;
            }
        }
        printf("%.3lf\n",res);
    }
    return 0;
}