PAT甲级真题 1079 Total Sales of Supply Chain (25分) C++实现(带深度的DFS)

博客介绍了PAT甲级考试中关于供应链总销售额问题的算法解决,采用C++的深度优先搜索(DFS)策略。在供应链网络中,每个成员以比供应商高r%的价格销售产品,零售商面对客户。通过DFS找到每个零售商的深度,计算其销售额,最终得出所有零售商的总销售额。

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题目

A supply chain is a network of retailers(零售商), distributors(经销商), and suppliers(供应商)– everyone involved in moving a product from supplier to customer.

Starting from one root supplier, everyone on the chain buys products from one’s supplier in a price P and sell or distribute them in a price that is r% higher than P. Only the retailers will face the customers. It is assumed that each member in the supply chain has exactly one supplier except the root supplier, and there is no supply cycle.

Now given a supply chain, you are supposed to tell the total sales from all the retailers.

Input Specification:

Each input file contains one test case. For each case, the first line contains three positive numbers: N (<=105), the total number of the members in the supply chain (and hence their ID’s are numbered from 0 to N-1, and the root supplier’s ID is 0); P, the unit price given by the root supplier; and r, the percentage rate of price increment for each distributor or retailer. Then N lines follow, each describes a distributor or retailer in the following format:

Ki ID[1] ID[2] … ID[Ki]

where in the i-th line, Ki is the total number of distributors or retailers who receive products from supplier i, and is then followed by the ID’s of these distributors or retailers. Kj being 0 means that the j-th member is a retailer, then instead the total amount of the product will be given after Kj. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the total sales we can expect from all the retailers, accurate up to 1 decimal place. It is guaranteed that the number will not exceed 1010.

Sample Input:
10 1.80 1.00
3 2 3 5
1 9
1 4
1 7
0 7
2 6 1
1 8
0 9
0 4
0 3
Sample Output:
42.4

思路

把销售渠道视为树,用邻接表e存储节点连接关系;零售商相当于叶子节点,用额外的数组num存储其货物量。

货物从根出发,每经过一层价格都会提高r%,经过n层的价格就是(1+r%) n

用dfs找出每个叶子节点的深度,进而求出它的价格,再乘以其货物量即可得到总销售额。由于图中无环,所以无需使用visited数组,dfs可以很简化。

小技巧:可先把原始售价视为1,到最后一步再乘以p。

代码

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

vector<vector<int> > e;
vector<int> num;
double n, p, r;
double sum = 0.0;

void dfs(int root, int layer){
    if (e[root].size()==0){
        sum += (num[root] * pow(r, layer));
        return;
    }
    for (int i=0; i<e[root].size(); i++){
        dfs(e[root][i], layer+1);
    }
}
int main(){
    cin >> n >> p >> r;
    e.resize(n);
    num.resize(n);
    r = (1 + r/100);
    for (int i=0; i<n; i++){
        int k;
        cin >> k;
        if (k==0){
            cin >> num[i];
        }
        while (k--){
            int temp;
            cin >> temp;
            e[i].push_back(temp);
        }
    }
    dfs(0, 0);
    printf("%.1lf\n", sum * p);
    return 0;
}

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