销售公寓递归定义

Question

我的任务如下：

一家建筑公司已完成一栋带有 A 公寓的新建筑 这些都可以在市场上买到。有 B 个潜在买家 这些公寓。

给定一个买家 b 和两个数组 s； p，值 s[b] 对应于 买家 b 想购买的公寓数量，p[b] 对应 买方愿意为公寓支付的金额。的目标 公司将确定它可以通过以下方式获得的最大利润 将公寓出售给合适的买家。请注意，并非所有 公寓需要出售，买方要么得到所有公寓 他要求或没有。

示例：假设有 8 套公寓待售和 4 套潜力 买家。每个买家想要购买的公寓数量和 他们愿意支付的价格在数组中可用：s = [4; 1个； 2； 3] 和 p = [8; 1个； 10个； 9]，分别。最大的保护 可以达到的是20。 3.1 [3 分] 假设 P(b;a) 是考虑到前 b 个买家的要求和当 a 公寓可供出售。给出 P(b;) 的递归定义 一）。

你们中有人知道如何递归解决这个问题吗？

谢谢

Answer 1

1.) 形成一个包含 (p/s) 值的数组，如 [2,2,2,2,1,5,5,3,3,3]。我在这里假设 (p/s) 的任何值都不会产生十进制数。

2.)现在，您可以将此问题简化为 k=8（待售公寓数量）的 K 组合问题。

3.) 打印出所有 k=8 且 n 为数组中的值的组合，您将看到使用此方法得到的最大值为 20。

Answer 2

认为这可能行得通。

class ApartmentBuyer{

    public static void main(String[] args){
        int noOfApartments=10;
        System.out.println("Program to find maximum profit from given set of buyers and apartments. \n total number of apartments="+noOfApartments);
        int [] buyersNeed= {4, 1, 2, 3};
        int [] price={8, 1, 10, 9};
        print(buyersNeed);
        print(price);
        System.out.println("Maximum Profit="+findMaxProfit(noOfApartments,buyersNeed,price,price.length-1,0));

    }

    /**
    *b={4} and p={1} and noOfApratment=4 , max profit=4
    *b={4,1} and price{1,5} noOfApartment= 1, max profit=4
    *Start from right most and go forward for recurence
    **/
    public static int findMaxProfit(int noOfApartments, int need[], int[] price, int pos, int profit){

        if(need==null || price==null || need.length<1 || price.length<1 || price.length!=need.length|| pos<0 || pos>need.length) return profit;
        if(noOfApartments<=0) return profit;
        int currentProfit=0;
        if(need[pos]<=noOfApartments) currentProfit= profit+need[pos]*price[pos];
        System.out.println("No Of Apartments:"+noOfApartments+ "\t pos:"+pos+"\t profit:"+profit+ "\tneed[pos]"+need[pos]);noOfApartments
        System.out.println("No Of Apartments:"+noOfApartments+ "\t pos:"+pos+"\t profit:"+profit+ "\tneed[pos]"+need[pos]);noOfApartments
        return Math.max(findMaxProfit(noOfApartments-need[pos],need,price, pos-1,currentProfit), findMaxProfit(noOfApartments,need,price,pos-1,profit));

    }


    static void print(int... arr){
        if(arr==null || arr.length<1) return;
        System.out.print("[");
        for(int i:arr)System.out.print(i+" ");
        System.out.println("]");
    }
}

Answer 3

#include <stdio.h>
#include <stddef.h>
#include <stdlib.h>

// Printing Matrix
void printdpmatrix(int l1, int l2, int dp[l1+1][l2+1]){
    for (int i= 0; i < l1+1; i++){
        printf("\n");
        for (int j = 0 ; j < l2+1; j++){
            printf("|%d|",dp[i][j]);
        }
    }
}
// Simple Max Function.
int max(int a, int b){return a > b ? a : b;}

// Returns maximum profit.
int maxprofit(int forsale, int* demand, int* price, int length){
    int ans;
    int sell;
    int notsell;
    //Declare Dp Matrix, +1 to accomodate the 0/0 basecase.  
    int dp[length+1][forsale+1];
    //Initialise matrix
    for (int i = 0; i < forsale+1; i++){dp[0][i] = 0;}
    for (int j = 0; j < length+1; j++){dp[j][0] = 0;}

    // Main logic loop 
    // i is the buyers 
    // j is the apartments
    for (int i = 1; i < length +  1; i++){
        for (int j = 1; j < forsale + 1; j++){
            //If we don't consider the current buyer we look at one buyer less. 
            notsell = dp[i-1][j];
            sell = 0; 
            //If the demand exceeds the supply, we can't sell anything. 
            if (demand[i-1] > j){ sell = 0;}
            //If we have the supply to cope with the demaind then we take
            //the profit of the current item and add the profit of this item 
            //disconsidered and with the supply reduced by the demand of this item. 
            else{ sell = price[i-1] + dp[i-1][j - (demand[i-1])];}
            //Update the DP matrix.
            dp[i][j] = max(sell,notsell);
            }
    }
    // Print matrix for fun. 
    printdpmatrix(length, forsale, dp);
    // Answer = last position in matrix. 
    ans = dp[length][forsale];
    // Print the answer for fun. 
    printf(" \n %d <-ANS\n",ans);
    return ans;
}
int main(){
    int houses = 8;
    int demand[4] = {4,1,2,3};
    int price[4] = {8,1,10,9};
    int length = 4; 
    maxprofit(houses, demand, price, length);

    return 0;
}

迟到总比不到好；) 来自信息 2 的问候。