[ACM_HDU_1009]FatMouse' Trade（贪心入门）

杰拉斯 | 时间：2012-04-18, Wed | 21,892 views
编程算法　

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22404 Accepted Submission(s): 6972

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

Sample Output


13.333
31.500

Source

HDU1009

其实这道题就相当于[ACM实验三]ACM程序设计基础（1）中可分割的背包问题，所谓贪心，就是每一次都找最大价值的作为当前阶段的结果。

应该注意的是float类型相除可能出现误差，而再乘以一个比较大数将使误差扩大，导致WA，因此要用double型。

代码如下：

#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
struct room{
	int j, f;
    bool operator < (const room t) const{
        return (float)j / f > (float)t.j / t.f;
    }

};
int main(){
	int m, n, i;
	while(scanf("%d%d", &m, &n) && (m + 1) || (n + 1)){
		vector<room> v;
		room r;
		double ans = 0;
		for(i = 0; i < n; ++i){
			scanf("%d%d", &r.j, &r.f);
			v.push_back(r);
		}
		sort(v.begin(), v.end());
		i = 0;
		while(m > 0 && i < v.size()){
			if(v[i].f <= m){
				m -= v[i].f;
				ans += v[i].j;
			}else{
				ans += (double)m / v[i].f * v[i].j;
				m = 0;
			}
			++i;
		}
		printf("%.3lf\n", ans);
	}
	return 0;
}

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