贪心——HDU题目1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33504    Accepted Submission(s): 10897

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 0
0 1
1 0
-1 -1

 

 

Sample Output

13.333
31.500
0.000
1.000

 

类似于0-1背包问题

#include <iostream>
#include <cstdio>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

typedef struct ROOM
{
    int j,f;
}ROOM;
ROOM room[10001];

bool cmp(ROOM a,ROOM b)
{
    return 1.0*a.j/a.f > 1.0*b.j/b.f;    
}
int main()
{
    int n,m,i;
    double max;
    while(scanf("%d%d",&m,&n) && (m!=-1||n!=-1))
    {
        if(n==0)//zhe里是特殊情况 
        {
            printf("0.000\n");continue;    
        }
        for(i=0;i<n;i++) 
            scanf("%d%d",&room[i].j,&room[i].f);    
        sort(room,room+n,cmp);
        //for(i=0;i<n;i++)
        //    printf("%d %d\n",room[i].j,room[i].f);
        max = i = 0;
        if(m==0)//第一次提交没考虑特殊情况
        {
            if(room[i].f==0)
            {
                max+=room[i].j;    
            }    
            i++;
        }
        while(m)
        {
            if(m>=room[i].f)
            {
                max += room[i].j;    
                m -= room[i].f;
            }
            else
            {
                max += 1.0 * room[i].j/room[i].f *m;    
                m=0;    
            }
            i++;
        }
        printf("%.3lf\n",max);
    }
    return 0;
}