zoukankan      html  css  js  c++  java
  • PAT1106:Lowest Price in Supply Chain

    1106. Lowest Price in Supply Chain (25)

    时间限制
    200 ms
    内存限制
    65536 kB
    代码长度限制
    16000 B
    判题程序
    Standard
    作者
    CHEN, Yue

    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 lowest price a customer can expect from some 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 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. All the numbers in a line are separated by a space.

    Output Specification:

    For each test case, print in one line the lowest price we can expect from some retailers, accurate up to 4 decimal places, and the number of retailers that sell at the lowest price. There must be one space between the two numbers. It is guaranteed that the all the prices will not exceed 1010.

    Sample Input:
    10 1.80 1.00
    3 2 3 5
    1 9
    1 4
    1 7
    0
    2 6 1
    1 8
    0
    0
    0
    
    Sample Output:
    1.8362 2

    思路
    类似 pat1079 的问题,bfs找到最长的路径然后计算价格就行。
    代码
    #include<iostream>
    #include<vector>
    #include<queue>
    #include<iomanip>
    using namespace std;
    vector<vector<int>> graph(100000);
    vector<bool> visits(100000,false);
    /*
    1.输入,构图
    2.bfs找最短
    3.迭代求乘积
    */
    
    
    int bfs(int& cnt)
    {
       queue<int> q;
       int minnum = 233333,level = 1,label = -1;
       q.push(0);q.push(label);
       while(!q.empty())
       {
         int tmp = q.front();
         q.pop();
         if(tmp == label)
         {
             if(q.empty())
                break;
             else
             {
                 level++;
                 q.push(label);
                 continue;
             }
         }
         visits[tmp] = true;
         if(graph[tmp].empty())
         {
             if(minnum > level)
             {
                 minnum = level;
                 cnt = 1;
             }
             else if (minnum == level)
             {
                 cnt++;
             }
         }
         else
         {
             for(int i = 0;i < graph[tmp].size();i++)
             {
                 if(!visits[graph[tmp][i]])
                    q.push(graph[tmp][i]);
             }
         }
       }
       return minnum - 1;
    }
    
    int main()
    {
        int n;
        double p,r;
        while(cin >> n >> p >> r)
        {
            r /= 100;
            for(int i = 0;i < n;i++)
            {
                int num;
                cin >> num;
                if(num == 0)
                    continue;
                for(int j = 0;j < num;j++)
                {
                    int tmp;
                    cin >> tmp;
                    graph[i].push_back(tmp);
                }
            }
            int cnt = 0;
            int len = bfs(cnt);
            for(int i = 0;i < len;i++)
            {
                p *= static_cast<double>(1 + r);
            }
            cout << fixed << setprecision(4) << p << " " << cnt << endl;
        }
    }
    

      

  • 相关阅读:
    管道
    linux基本网络配置
    201671010130 201620172 《Java程序设计》第三周学习小结
    201671010130 201620172 《Java程序设计》第二周学习小结
    201671010130 201620172 《Java程序设计》第五周学习小结
    201671010130 201620172 《Java程序设计》首次与Java打交道
    朵的面向对象程序设计课程学习进度条
    201671010130 201620172 《Java程序设计》第六七周学习小结
    201671010130 201620172 《Java程序设计》第四周学习小结
    Note of Effective Modern C++: 42 Specific Ways to Improve Your Use of C++11 and C++14
  • 原文地址:https://www.cnblogs.com/0kk470/p/8004897.html
Copyright © 2011-2022 走看看