hdu1009 优先队列

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33106    Accepted Submission(s): 10731

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

 

#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>
using namespace std;
struct node
{
	friend bool operator< (node n1,node n2)      //在这个分函数定义中“operator<”是固定的名称，用来定义优先级如何比较，不能改用其他名字
	{
		double a,b;
		a=n1.a*1.0/n1.b;
		b=n2.a*1.0/n2.b;
		return a<b;                          //这里是来定义优先级如何比较，是从大到小还是别的，这里定义的是从大到小
	}
	int a,b;
};
int main()
{
	int m,n,i;
	double max;
	while(scanf("%d%d",&m,&n)>0)
	{
		if(m==-1&&n==-1) break;max=0.0;
		priority_queue<node> qi;                 //在main函数中定义优先队列
		node s[10000];
		for(i=1;i<=n;i++)
			scanf("%d%d",&s[i].a,&s[i].b);
		for(i=1;i<=n;i++)
			qi.push(s[i]);                   //将元素压入到队列中
		for(i=1;i<=n;i++)
		{
			if(qi.top().b<=m)
			{
				m-=qi.top().b;           //取出最顶端元素，也就是最优结果
				max+=qi.top().a;
			}
			else
			{
				double j;
				j=m*1.0/qi.top().b;m=0;
				max+=qi.top().a*j;
			}
			if(m==0) break;
			qi.pop();                        //去掉顶端
		}
		printf("%.3lf\n",max);
	}
	return 0;
}