[ACM_HDU_1009]FatMouse' Trade（贪心算法入门）

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
5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

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; } }; //_hdu_1009 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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