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[ACM_HDU_1009]FatMouse' Trade（贪心入门）

杰拉斯 杰拉斯 | 时间：2012-04-18, Wed | 22,734 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

  1.  
  2. 5 3
  3. 7 2
  4. 4 3
  5. 5 2
  6. 20 3
  7. 25 18
  8. 24 15
  9. 15 10
  10. -1 -1

Sample Output

  1.  
  2. 13.333
  3. 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：

  1. #include<stdio.h>
  2. #include<vector>
  3. #include<algorithm>
  4. using namespace std;
  5. struct room{
  6. 	int j, f;
  7.     bool operator < (const room t) const{
  8.         return (float)j / f > (float)t.j / t.f;
  9.     }
  10.  
  11. };
  12. int main(){
  13. 	int m, n, i;
  14. 	while(scanf("%d%d", &m, &n) && (m + 1) || (n + 1)){
  15. 		vector<room> v;
  16. 		room r;
  17. 		double ans = 0;
  18. 		for(i = 0; i < n; ++i){
  19. 			scanf("%d%d", &r.j, &r.f);
  20. 			v.push_back(r);
  21. 		}
  22. 		sort(v.begin(), v.end());
  23. 		i = 0;
  24. 		while(m > 0 && i < v.size()){
  25. 			if(v[i].f <= m){
  26. 				m -= v[i].f;
  27. 				ans += v[i].j;
  28. 			}else{
  29. 				ans += (double)m / v[i].f * v[i].j;
  30. 				m = 0;
  31. 			}
  32. 			++i;
  33. 		}
  34. 		printf("%.3lf\n", ans);
  35. 	}
  36. 	return 0;
  37. }

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