贪心算法D 猫和老鼠c++

题目描述:

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:老鼠有M磅猫的食物,有N个房间,每个房间有J[i]的豆子,猫能吃F[i]食物,当老鼠给猫a%的食物,就能获得a%的豆子。问老鼠最多能得到多少豆子。

思路:

这个题目是贪心算法的常规问题,算法和背包问题很像(不是0/1背包问题)。 

首先将每个房间的豆子和猫食的比按递增排序。(类似于背包问题的单位价值比)

然后按照这个排序,尽量在不超过M的情况下,首先选择豆子比最高的房间。

最后稍微处理一下最后一个房间(可能会超过M)

 

代码:

#include
#include
#include
#define N 1005
using namespace std;

struct room{
    double javabean;
    double food;
}r[N];

bool cmp(room A, room B){    //递增排序,按照房间豆子比
    if((A.javabean/A.food) > (B.javabean/B.food)) return true;
    else return false;
}

double maxJavabeans(room *r,int m,int n)
{
    double sum=0;
    int i=0;
    while(m>0&&i>m>>n&&m!=-1&&n!=-1){
        for(int i=0;i>r[i].javabean>>r[i].food;
        }
        sort(r,r+n,cmp);
        cout<

 

 

 

 


 

 

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