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HDU1009 FatMouse' Trade 【贪心】

2024-07-14 16:44:02   223人阅读

FatMouse‘ Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42786    Accepted Submission(s): 14274


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 37 24 35 220 325 1824 1515 10-1 -1
 


 

Sample Output
13.33331.500

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 1002using 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;}


 


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