PAT1090. Highest price in supply Chain

A supply chain is a network of retailers (retailer), distributors (dealer), and suppliers (supplier)--everyone involved in moving a produ CT 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 Distribu   Te them in a price, is r% higher than P. It is assumed this member in the supply chain have exactly one supplier except the root supplier, and there is no supp Ly cycle.

Now given a supply chain and you is supposed to tell the highest price we can expect from some retailers.

Input Specification:

Each input file contains the 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 is numbered from 0 to N-1); P, the price of 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 and each number Si are the index of the supplier for the i-th member.  Sroot for the root supplier are defined to be-1. All the numbers in a line is separated by a space.

Output Specification:

For all test case, print on 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 numbers. It is guaranteed, the price would not exceed 1010.

Sample Input:

9 1.80 1.001 5 4 4-1 4 5 3 6

Sample Output:

1.85 2

1:print in one line the highest price we can expect from some retailers: from  some retailers.
2:the number of retailers that sell at the highest.       Highest. That means highest depth.
3:  sroot for the root supplier are defined to be-1.      That means the value corresponding to I is root.

1#include <iostream>2#include <cstdio>3#include <vector>4#include <cmath>5 using namespacestd;6 #defineMAX 1000107 intN;8 Doublep,r,ans=0;9 intMaxdepth=0;Ten intnum=0; One structNode A { -vector<int>Child ; - }node[max]; the voidDFS (intIndexintdepth) - { -     if(node[index].child.size () = =0) -     { +         if(depth>maxdepth) -         { +Maxdepth=depth; Anum=1; at         } -         Else if(depth==maxdepth) -         { -num++; -         } -         return; in     } -      for(intI=0; I<node[index].child.size (); i++) to     { +DFS (node[index].child[i],depth+1); -     } the } *  $ Panax Notoginseng intMainintargcChar*argv[]) - { the     introot=max-1; +scanf"%D%LF%LF",&n,&p,&R); AR/= -; the      for(intI=0; i<n;i++) +     { -         inttem; $scanf"%d",&tem); $         if(tem==-1) -root=i; -         Else the Node[tem].child.push_back (i); -     }WuyiDFS (Root,0); theprintf"%.2LF%d\n", P*pow (1+r,maxdepth), num); -      Wu     return 0; -}

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PAT1090. Highest price in supply Chain