FatMouse‘ Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem 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

题意：给定一个容量为m的背包以及n个物品和每个物品的重量及价值，单个物品可以任意切分，求背包能获得的最大装载价值。

题解：可以求出每个物品的单位重量价值，排序，每次选择单位价值最大的装包即可。

#include <stdio.h>
#include <algorithm>
#define maxn 1002
using std::sort;

struct Node{
	int v, c;
	double val;
} arr[maxn];

bool cmp(Node a, Node b)
{
	return a.val > b.val;
}

int main()
{
	//freopen("in.txt", "r", stdin);
	//freopen("out.txt", "w", stdout);
	int m, n, i, a, b;
	double ans;
	while(scanf("%d%d", &m, &n), m >= 0 && n >= 0){
		for(i = 0; i < n; ++i){
			scanf("%d%d", &a, &b);
			arr[i].val = a * 1.0 / b;
			arr[i].v = a; arr[i].c = b;
		}
		ans = 0;
		sort(arr, arr + n, cmp);
		for(i = 0; i < n; ++i){
			if(m >= arr[i].c){
				m -= arr[i].c; ans += arr[i].v;
			}else{
				ans += m * arr[i].val; break;
			}
		}
		printf("%.3lf\n", ans);
	}
	return 0;
}

HDU1009 FatMouse' Trade 【贪心】