算法笔记练习 4.4 贪心 问题 E: FatMouse's Trade

算法笔记练习 题解合集

题目链接

题目

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

4 2
4 7
1 3
5 5
4 8
3 8
1 2
2 5
2 4
-1 -1

样例输出

2.286
2.500

思路

类似这个题：PAT B1020 月饼

要用有限的钱（cat food）去换到尽量多的东西（javabeans），那么要尽可能地买便宜的东西，也就是单价低的东西。

所以将所有货仓按照单价从低到高排序，从前往后尽可能地买。

代码

#include 
#include 
typedef struct{
	double java; // 有多少货
	double food; // 全部买下需要的钱
	double avg;  // 单价
} Warehouse;
// qsort 排序函数 
int cmpAvg(const void *ca, const void *cb){
	return (*(Warehouse*)ca).avg > (*(Warehouse*)cb).avg;
} 
int main(){
	double M;
	int N, i;
	while (scanf("%lf %d", &M, &N) != EOF){
		if (N == -1) continue;
		// 输入数据，排序 
		double ans = 0;
		Warehouse ware[N]; 
		for (i = 0; i < N; ++i){
			scanf("%lf %lf", &ware[i].java, &ware[i].food);
			ware[i].avg = ware[i].food / ware[i].java;
		}
		qsort(ware, N, sizeof(Warehouse), cmpAvg);
		// 从单价低的货仓开始买
		// 直到全部买完或者不足以买下一个完整的货仓为止 
		for (i = 0; i < N; ++i){
			if (ware[i].food < M){
				M -= ware[i].food;
				ans += ware[i].java; 
			} else break;
		}
		// i < N 说明没有全部买完 
		if (i < N)
			ans += ware[i].java * M / ware[i].food;
		printf("%.3lf\n", ans);
	} 
	return 0;
}