【ACM】HDOJ 1009 FatMouse' Trade

HDOJ 1009

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.

题目描述：

胖老鼠准备了M榜的猫食，想要用来与守卫藏有他最爱的JavaBean的仓库的猫猫们进行交易。

仓库有N个房间，第i个房间里面有J[i]榜的JavaBean，同时需要F[i]榜的猫食才能换取。

胖老鼠可以只按比例换取JavaBean而不是换取整个房间的JavaBean。

现在它把交易计划交给你来处理：告诉它，它最多能得到多少榜JavaBean。

输入说明：

允许多次案例输入，每个案例的第一行为两个无符号整数 M和N。

接着是N行输入，每一行包含无符号整数J[i]和F[i]。

程序输入以两个-1结束。

所有的整数都不会超过1000。

输出说明：

对于每个案例，输出一个保留三位有效数字的实数，用来表示这个案例中胖老鼠最大能获得的JavaBean。

我的思路：

要使JavaBean最大，那就得要每一榜猫食换取尽可能多的JavaBean。

所以 先计算出每一个房间的 单位食物能换取的JavaBean，即 V[i] = J[i]/F[i]。

接着 对数组V排序，V[i]越大，猫食用的越值得。

剩余的猫食小到不足以换取整个房间的JavaBean时，选择按比例换取JavaBean。

#include <iostream>
#include <iomanip>
//#include <fstream>
#include <cstdlib>
#include<algorithm>

using namespace std;
struct s_ROOM
{
    double JB;
    double food;
    double value;

}Rooms[1000];
double GetJavaBean(int);
bool cmpValue(s_ROOM,s_ROOM);

int M,N;
int main()
{
 //   ifstream input("F:\input.txt");

    while(cin>>M>>N && M!=-1 && N!=-1)
    {

        for(int i=0;i<N;i++)
        {
            cin>>Rooms[i].JB>>Rooms[i].food;
            Rooms[i].value  = Rooms[i].JB/Rooms[i].food;
        }
        sort(Rooms,Rooms+N,cmpValue);
        cout<<fixed;
        cout<<setprecision (3)<<GetJavaBean(M)<<endl;
    }
    return 0;
}

double GetJavaBean(int totalFood)
{
    double ans=0;
    int leftFood = totalFood;
    int i=0;
    for(i=0;i<N && leftFood>= Rooms[i].food;i++)
    {
    //    cout<<Rooms[i].JB<<','<<Rooms[i].food<<','<<Rooms[i].value<<endl;
        ans+=Rooms[i].JB;
        leftFood-=Rooms[i].food;
    }
    if(i<N)
    {
      //  cout<< Rooms[i].JB*leftFood/Rooms[i].food<<endl;
        ans+=Rooms[i].JB*leftFood/Rooms[i].food;
    }



    return ans;
}

bool cmpValue(s_ROOM r1,s_ROOM r2)
{

    return r1.value-r2.value >0;
}

其中用到了STL（标准模板库）中的sort函数用来对value进行大到小排序。

AC要点：实数类型选择double！