南阳oj贪心算法之Greedy Mouse

最新推荐文章于 2022-05-09 15:15:06 发布

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#贪心算法

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探讨了在有限猫粮条件下，贪心小鼠如何通过最优策略从守卫花生仓库的猫咪那里换取最大数量的花生。算法通过比较各仓库花生与猫粮交换比率，优先选择最有利的交易，实现资源最大化利用。

/**
Greedy Mouse
时间限制：1000 ms  |  内存限制：65535 KB
难度：3
描述
A fat mouse prepared M pounds of cat food,ready to trade with the cats guarding the warehouse containing his

favorite food:peanut. The warehouse has N rooms.The ith room containsW[i] pounds of peanut and requires 

F[i] pounds of cat food. Fatmouse does not have to trade for all the peanut in the room,instead,he may get 

 W[i]*a% pounds of peanut if he pays F[i]*a% pounds of cat food.The mouse is a stupid mouse,so can you tell 

him the maximum amount of peanut he can obtain.

输入
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 W[i] and F[i] respectively. The test case is terminated by two -1.
All integers are not greater than 1000.
输出
For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of penaut that FatMouse can obtain.
样例输入
5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1
样例输出
13.333
31.500
上传者
*/
#include<iostream>
#include<algorithm>
#define maxsize 1000
using namespace std;
struct node{
	int w;
	int v;
	double r;
};
int cmp(node a,node b)
{
	return a.r>b.r;
}
int main()
{
	int m, n;
	while(cin>>m>>n)
	{
		if(m==-1&&n==-1)
			break;
		node s[maxsize];
		int i,j;
		for(i=0;i<n;i++)
		{
			cin>>s[i].w>>s[i].v;
			s[i].r=(s[i].w)*1.0/s[i].v; 
		}
		sort(s,s+n,cmp);
		double sum=0;
		for(i=0;m>0&&i<n;i++)
		{
			if(s[i].v<m)
			{
				sum+=s[i].w;
				m-=s[i].v;
			}
			else
			{
				sum+=m*s[i].r;
				break;
			}
		}
		printf("%.3f\n",sum);
	}
	return 0;
}