HDU 1009 FatMouse' Trade（簡單貪心，乍一看以爲是0-1揹包問題，差點想多了）

題目鏈接

FatMouse’ Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 116539 Accepted Submission(s): 40213

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

思路：
定義一個性價比，從大到小排序，貪心策略，把最值錢的儘量拿完。

完整代碼

#include <iostream>
#include <algorithm>
#include <iomanip>

using namespace std;

struct room{
    double j,f;
    double oprice;
    friend bool operator <(room a,room b){
        return a.oprice>b.oprice;
    }
}Room[1005];

int main()
{
    int n,m;
    while(cin>>n>>m){
        if(n==-1&&m==-1)
            break;
        for(int i=0;i<m;i++){
            cin>>Room[i].j>>Room[i].f;
            Room[i].oprice=Room[i].j/Room[i].f;
        }
        sort(Room,Room+m);

        double sum=0.0;
        for(int i=0;i<m;i++){
            if(n>=Room[i].f){
                n-=Room[i].f;
                sum+=Room[i].j;
            }
            else{
                sum+=Room[i].oprice*n;
                break;
            }
        }
        cout<<setiosflags(ios::fixed)<<setprecision(3)<<sum<<endl;
    }
    return 0;
}