HDU1009 贪心思想

FatMouse' Trade
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
Submit  Status  Practice  HDU 1009

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
 

这个题的大意是:它有M猫食,N个房间,每个房间有f[i]猫食,j[i]量的 javabean,按一定的比例拿猫食来保护javabean。问一共有M猫食最多能保护多少javabean。

简单的贪心,只需按每个房间的javabean和猫食的比例从高到底排序,然后贪心即可。

代码中我处理了输入时f[i]等于0的情况,但交上去WA,我删去之后就AC了,看来没必要处理啊,,


#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;

struct food
{
	int f,j;
	double ave;
}s[1010];

int cmp(const food &a,const food &b)
{
	return a.ave>b.ave;
}

int main()
{
	int i,k,n,m;
	double ans;
	while(scanf("%d%d",&m,&n))
	{
		if(m==-1 && n==-1)
			break;
		for(i=0;i<n;i++)
		{
			scanf("%d%d",&s[i].j,&s[i].f);
			s[i].ave=(double)s[i].j/s[i].f;
		}
		sort(s,s+n,cmp);
		ans=0;
		for(i=0;i<n;i++)
		{
			if(s[i].f<=m)
			{
				ans+=(double)s[i].j;
				m-=s[i].f;
			}
			else
			{
				ans+=s[i].ave*m;
				break;
			}
		}
		printf("%.3f\n",ans);
	}
	return 0;
}


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