• HDU-1009FatMouse' Trade


    FatMouse' Trade

     

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

     

    题目链接

    http://acm.hdu.edu.cn/showproblem.php?pid=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. 
     

     

    Sample Input
    5 3
    7 2
    4 3
    5 2
    20 3
    25 18
    24 15
    15 10
    -1 -1
     

     

    Sample Output
    13.333
    31.50
     
    总结
     

    初步了解了贪心算法,是选择价值比最大的东西来拿(由于可以切开拿,不是必须整体),接下来就好多了。

    遇到的问题主要是浮点数的误差,只能用价值比(浮点数)比较大小,后面的运算求和时不能用它相乘来计算,拿样例来说,结果是13.333,而我没改进前,算出来是12.444.可见是误差。

    第二个就是网上其他人使用结构体来储存,这样排序时其他变量能一起移动,但我开了3个数组,而且也不能用sort来排序了,只能自己写了选排。

    #include<iostream>
    #include<cstdio>
    using namespace std;
    double price[1005];
    int num1[1005],num2[1005];
    
    void sort1(double* a,int x)
    {
        int b;
        double c;
        for(int i=0;i<x;i++){
            for(int j=i+1;j<x;j++){
                if(a[i]<a[j]){
                    c=a[i];
                    a[i]=a[j];
                    a[j]=c;
                    b=num1[i];
                    num1[i]=num1[j];
                    num1[j]=b;
                    b=num2[i];
                    num2[i]=num2[j];
                    num2[j]=b;
                }
            }
        }
    }
    
    int main()
    {
        int m,n,j;
        while(scanf("%d %d",&m,&n)==2){
            if(m==-1&&n==-1) break;
            for(int i=0;i<n;i++){
                scanf("%d %d",&num2[i],&num1[i]);
                price[i]=(double)num2[i]/num1[i];
            }
            sort1(price,n);
            //for(int i=0;i<n;i++) cout <<price[i]<<num[i]<<endl;
            double sum=0;
            for(int i=0;i<n;i++){
                if(m<=0) break;
                if(m>=num1[i]){
                    sum+=num2[i];
                    m-=num1[i];
                }
                else {
                    sum+=(double)m/num1[i]*num2[i];
                    m=0;
                }
            }
            printf("%.3lf
    ",sum);
        }
        return 0;
    }

     

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  • 原文地址:https://www.cnblogs.com/wangdongkai/p/5267451.html
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