FatMouse’ Trade

╰半橙微兮° 2022-04-24 06:20 141阅读 0赞

## 简介 ##

贪心算法（又称贪婪算法）是指，在对问题求解时，总是做出在当前看来是最好的选择。也就是说，不从整体最优上加以考虑，他所做出的是在某种意义上的局部最优解。  
贪心算法不是对所有问题都能得到整体最优解，关键是贪心策略的选择，选择的贪心策略必须具备无后效性，即某个状态以前的过程不会影响以后的状态，只与当前状态有关。

## FatMouse’ Trade ##

### FatMouse’ Trade ###

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 33703 Accepted Submission(s): 10981

### 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

### 贪心思想 ###

从每傍食物最便宜的开始

### C++代码 ###

#include<iostream>
    #include<algorithm>
     
    using namespace std;
     
    struct bean{
        double pound;
        double value;
        double perv;
        bool operator < (const bean &A)const{
            return perv > A.perv;
        }
    }buff[100];
     
    int main(){
        int n;
        double m;
        while(cin>>m>>n){
            if(m == -1 && n == -1)break;
            for(int i = 0; i < n; i++){
                scanf("%lf%lf", &buff[i].pound, &buff[i].value);
                buff[i].perv = buff[i].pound/buff[i].value;
            }
            sort(buff, buff + n);
            int id = 0;
            double ans = 0;
            while(m > 0 && id < n){
                if(m > buff[id].value){
                    m -= buff[id].value;
                    ans += buff[id].pound;
                }else{
                    ans += m*buff[id].perv;
                    m = 0;
                }
                id++;
            }
            cout<<ans<<endl;
        }
        return 0;
    }
    /*
    5 3
    7 2
    4 3
    5 2
    20 3
    25 18
    24 15
    15 10
    -1 -1
    */