zoukankan      html  css  js  c++  java
  • 1079 Total Sales of Supply Chain (25 分)

    1079 Total Sales of Supply Chain (25 分)

    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 N1, 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
    #include<iostream>
    #include<vector>
    #include<algorithm>
    #include<queue>
    #include<string>
    #include<map>
    #include<set>
    #include<stack>
    #include<string.h>
    #include<cstdio>
    #include <unordered_map>
    #include<cmath>
    
    using namespace std;
    vector<vector<int>> graph;
    int units[100005];
    double sum=0;
    int n;
    double p,r;
    
    void dfs(int root,int depth)
    {
        if(graph[root].size()==0)
        {
            //cout<<p<<" "<<pow(1+r/100,depth)<<" "<<units[root]<<endl;
            sum+=p*pow(1+r/100,depth)*units[root];
        }
        for(auto num:graph[root])
            dfs(num,depth+1);
    }
    int main()
    {
    
        scanf("%d%lf%lf",&n,&p,&r);
        graph.resize(n);
        for(int i=0;i<n;i++)
        {
            int k,unit,id;
            scanf("%d",&k);
            if(k==0)
            {
                scanf("%d",&unit);
                units[i]=unit;
                continue;
            }
            for(int j=0;j<k;j++)
            {
                scanf("%d",&id);
                graph[i].push_back(id);
            }
        }
        dfs(0,0);
        printf("%.1f",sum);
        return 0;
    }
  • 相关阅读:
    sql server 删除重复数据新思路
    sqlserver 迁移 mysql
    ASP.NET Web deployment task failed. 请与服务器管理员联系,检查授权和委派设置 部署任务失败的解决方案
    数据库交互之减少IO次数
    sqlserver 安全设置
    windows设置相对路径的快捷方式
    利用SignalR实现实时推送信息
    image magick 备忘
    dotnetCore开发中遇到的一些问题
    “NETSDK1061: 项目是使用 Microsoft.NETCore.App 版本 2.1.14 还原的, 但使用当前设置, 将改用版本 2.1.0。”的处理方法
  • 原文地址:https://www.cnblogs.com/zhanghaijie/p/10329952.html
Copyright © 2011-2022 走看看