HDU 1009 FatMouse' Trade

最新推荐文章于 2021-05-09 21:32:11 发布

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本文介绍了一道关于FatMouse利用有限猫粮交换仓库中JavaBeans的最大价值的问题。通过贪心算法，优先选择性价比高的交易方案，实现最大化的利益获取。

FatMouse’ Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 76179 Accepted Submission(s): 26096

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

Author
CHEN, Yue

题意

意思是有一个老鼠想吃食物，但是食物被猫看守，老鼠可以用诱饵引诱猫，这样就能去吃食物。不同食物需要消耗的诱饵不一样，价值也不一样。求最多能获得多大的价值。
（最奇葩的是，如果用3个诱饵能完全吃完食物，你用1个诱饵就能吃1/3个食物。！！！，不过好像也很科学哎。）

这就是个贪心问题，我们肯定要优先选择最高性价比的食物。

解题思路

首先按照性价比排序，优先选择最高性价比的食物。
定义一个 double 型变量 sum=0，计算能获得食物的价值，定义一个 w 统计剩余的诱饵。
查看自己所剩余的诱饵够不够完全吃完食物，如果能，就扔出所需要的诱饵，如果不能就全部扔出去。然后sum=sum+这次获得的食物价值。

AC代码：

#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;

struct sub
{
    int v,w;
    double percent;
};

int w,n;
double sum=0;
sub arr[2003];

bool cmp(sub a,sub b){
    return a.percent>b.percent;
}

void f(sub a){
    if(w>=a.w){
        w-=a.w;
        sum+=a.v;
    }
    else{
        sum+=w*a.percent;
        w=0;
    }
    //printf("test case is %.3lf and has w=%d\n",sum,w);
}

int main(){
    while(~scanf("%d %d",&w,&n)&&w!=-1){
        sum=0;
        for(int i=0;i<n;++i){
            scanf("%d %d",&arr[i].v,&arr[i].w);
            arr[i].percent=arr[i].v*1.0/arr[i].w;
        }
        sort(arr,arr+n,cmp);
        for(int i=0;i<n;++i)
            f(arr[i]);
        printf("%.3lf\n",sum);
    }
}