HDU——1009FatMouse' Trade（贪心+结构体+排序）

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 60654    Accepted Submission(s): 20406

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

题目大意：（转载）老鼠有M磅猫食。有N个房间，每个房间前有一只猫，房间里有老鼠最喜欢的食品JavaBean,J[i]。若要引开猫，必须付出相应的猫食F[i]。当然这只老鼠没必要每次都付出所有的F[i]。若它付出F[i]的a%，则得到J[i]的a%。求老鼠能吃到的做多的JavaBean。

代码：

#include<iostream>
#include<string>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;
struct food
{
	double ji;
	double fi;
	double xingjiabi;//按照性价比排序
};
bool cmp(const food &a,const food &b)
{
	if(a.xingjiabi!=b.xingjiabi)
		return a.xingjiabi>b.xingjiabi;//性价比高的排前面
	else
		return a.fi<b.fi;//性价比一样则需要代价的fi（猫粮）小的排前面，不过再看看我后面的代码似乎不需要这条
}
int main(void)
{
	int m,n;
	double ans,rest;
	while (cin>>m>>n)
	{
		if(m==-1&&n==-1)
			break;
		food *list=new food[n];
		for (int i=0; i<n; i++)
		{
			scanf("%lf %lf",&list[i].ji,&list[i].fi);
			list[i].xingjiabi=list[i].ji*1.0/list[i].fi;
		}
		sort(list,list+n,cmp);
		ans=0;
		rest=m;
		for (int i=0; i<n; i++)
		{
			if(rest>=list[i].fi)//可以全部拿到直接相加
			{
				rest-=list[i].fi;
				ans+=list[i].ji;
			}
			else
			{
				ans=ans+list[i].xingjiabi*rest;//只能拿到一部分则把剩下的部分全给了，按照付出的百分比进行加成
				rest-=rest;
			}
		}
		printf("%.3lf\n",ans);
		delete []list;
	}
	return 0;
}