贪心初步-FatMouse' Trade

这道题是真的很简单，只需要用优先队列以单位价值作为排序规则就可以解决。

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 46804    Accepted Submission(s): 15704

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

 

Author

CHEN, Yue

 

Source

ZJCPC2004

 

Recommend

JGShining

 

附上代码：

#include <cstdio>
#include<queue>
#define maxn 1005;
using namespace std;

struct Food{
	int hav,ned;
	double va;
};

bool operator < (struct Food a,struct Food b){
	if(a.va < b.va)
		return	true;
	else
		return	false;
}
int main(){
	int m,n;
	struct Food	ha[1005];
	while(scanf("%d %d",&m,&n) && m!=-1&&n!=-1){
		priority_queue<struct Food> lis;
		for(int i = 0;i < n;i++){
			scanf("%d %d",&ha[i].hav,&ha[i].ned);
		}
		for(int i = 0;i < n;i++){
			ha[i].va = ha[i].hav*1.0/ha[i].ned;
			lis.push(ha[i]);
		}
		double cnt = 0.0;
		while(!lis.empty() && m){
			struct	Food t;
			t = lis.top();
			if(m >= t.ned){
				m -= t.ned;
				cnt += t.hav;
				lis.pop();
			}else{
				cnt += t.va * m;
				m = 0;
				lis.pop();
			}
		}
		printf("%.3f
",cnt);
	}
	return	0;
}