PAT1090:Highest Price in Supply Chain

Posted 2020-10-09 0kk470

1090. Highest 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. 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 highest price we 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 they are numbered from 0 to N-1); P, the price given by the root supplier; and r, the percentage rate of price increment for each distributor or retailer. Then the next line contains N numbers, each number S_i is the index of the supplier for the i-th member. S_root for the root supplier is defined to be -1. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the highest price we can expect from some retailers, accurate up to 2 decimal places, and the number of retailers that sell at the highest price. There must be one space between the two numbers. It is guaranteed that the price will not exceed 10¹⁰.

Sample Input:

9 1.80 1.00
1 5 4 4 -1 4 5 3 6

Sample Output:

1.85 2

思路
这道题归根结底就是用dfs求图（或者说树）的根节点到终点（或者说树的叶节点）最大路径maxLength，那么利润的增长次数就是maxLength - 1，此时价格最大。那么步骤可以是：

1.根据输入构造图
2.dfs找最大路径maxlength，并统计最大路径的次数。
3.最大价格 = price * pow( 1 + rate , maxLength - 1)
4.输出最大价格（两位小数）和最大路径的出现次数。

代码

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

vector<int> levels(100002,0);


//dfs and count level‘s num
void dfs(const vector<vector<int>>& graph,int root,int level)
{
   levels[level]++;
   if(graph[root].empty())
      return;
   for(int i = 0;i < graph[root].size();i++)
     dfs(graph,graph[root][i],level + 1);
}


int main()
{
  int N;
  double price,rate;
  while(cin >> N >> price >> rate)
  {
      vector<vector<int>> graph(N);
      int root = 0;
      //build graph
      for(int i = 0;i < N;i++)
      {
          int father;
          cin >> father;
          if(father == - 1)
            root = i;
          else
            graph[father].push_back(i);
      }
      
      //dfs
      dfs(graph,root,1);
      
      //find the maximum depth
      int maxLength = 0;
      for(int i = 1; i <levels.size();i++)
      {
          if(levels[i] == 0)
          {
              maxLength = i - 1 ;
              break;
          }
      }
      //calculate price
      for(int i = 0;i < maxLength - 1;i++)
      {
          price *= (1 + rate/100);
      }
      cout << fixed << setprecision(2) << price << " ";
      cout << levels[maxLength] << endl;
      levels.clear();
  }
}