hdu-1009\tFatMouse\' Trade（简单贪心）

最新推荐文章于 2018-10-03 22:45:43 发布

Nature_Ran

最新推荐文章于 2018-10-03 22:45:43 发布

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分类专栏：贪心算法题

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本文介绍了一种算法问题，即如何帮助FatMouse通过最优策略用猫粮交换到最多的Java豆。该问题涉及到多个房间，每个房间包含不同数量的Java豆和所需的猫粮。通过排序和贪心策略，确定了FatMouse能获得的最大Java豆数量。

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FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 53691 Accepted Submission(s): 17943

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

Sample Output

   
    13.333
31.500

原题

#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
struct node{
    int j,f;
    double one;
}room[1111];
int cmp(node a,node b){
    return a.one>b.one; 
}
int main(int argc, char *argv[])
{
    int n,m,i;
    while(cin>>m>>n){
        if(n==-1&&m==-1) break;
        for(i=0;i<n;i++){
            cin>>room[i].j>>room[i].f;
            room[i].one=1.0*room[i].j/room[i].f;
        }
        sort(room,room+n,cmp);
        double sum=0;
        for(i=0;i<n;i++){
            if(m>=room[i].f){
                sum+=room[i].j;
                m-=room[i].f;
            }
            else {
                sum+=room[i].one*m;
                break;
            }
        }
        printf("%.3f\n",sum);
    }
    return 0;
}