九度OJ 1433 贪心算法

最新推荐文章于 2021-06-29 10:43:34 发布

码农小非

最新推荐文章于 2021-06-29 10:43:34 发布

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分类专栏：数据结构与算法

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本博客介绍了一个利用贪心算法解决FatMouse在仓库中交易JavaBeans和猫粮的问题。通过每次选择性价比最高的交易，直至资金用尽，最大化获得的JavaBeans数量。详细解释了算法实现和输入输出格式，提供了样例输入输出帮助理解。

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题目链接：http://ac.jobdu.com/problem.php?pid=1433

题目描述：

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.

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 J[i] and F[i] respectively. The last test case is followed by two -1's. 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 JavaBeans that FatMouse can obtain.

样例输入：

样例输出：

13.333
31.500

所谓贪心算法说得最简单就是每次都取当前最好的情况。这题就是简单的贪心算法实例，每次都取性价比高的买，直到没钱了为止。

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

struct Food{
	double quality;
	double value;
	double ratio;
}buf[1000];

bool cmp(Food a,Food b){
	return a.ratio>b.ratio;
}

int main(){
	int n;
	double m;
	while(scanf("%lf%d", &m,&n)!=EOF){
		if(m==-1&&n==-1)
			break;
		for(int i=0;i<n;i++){
			scanf("%lf%lf",&buf[i].quality,&buf[i].value);
			buf[i].ratio=buf[i].quality/buf[i].value;
		}
		double recFood=0;
		sort(buf,buf+n,cmp);
		for(int i=0;i<n;i++){
			if(m>=buf[i].value){
				recFood+=buf[i].quality;
				m-=buf[i].value;
			} else{
				recFood+=buf[i].ratio*m;
				break;
			}
		}
		printf("%.3lf\n",recFood);
	}
	return 0;
}