九度OJ 1433 贪心算法

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：

5 37 24 35 220 325 1824 1515 10-1 -1

样例输出：

13.33331.500

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;}