A supply chain is a network of retailers (retailer), distributors (reseller), and suppliers (suppliers)--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 and is r% higher than P. Only the retailers would face the customers. It is assumed this supply chain has exactly one supplier except the root supplier, and there is no supp Ly cycle.
Now given a supply chain, your are supposed to tell the lowest price a customer can expect from some retailers.
Input Specification:
Each input file contains one test case. For the 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 to 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, the "Ki is" total number of distributors or retailers who receive all products from supplier I, and is T Hen followed by the ID ' s of these distributors or retailers. Kj being 0 means the j-th member is a retailer. All of the numbers in a line are separated by a.
Output Specification:
For each test case, print at one line the lowest price we can expect from some retailers, accurate up to 4 decimal places, And the number of retailers that sell in the lowest price. There must is one space between the two numbers. It is guaranteed the "all" prices won't exceed 1010. Sample Input:
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
Recently did the topic of the binary tree become addicted.
Key point: Build, find out the depth of the most shallow leaf node and its number can be.
#include <iostream>
#include <string.h>
#include <stack>
#include <vector>
using namespace std;
#include <stdio.h>
#include <algorithm>
#include <queue>
vector<int>tree[ 150000];
int n,mm,ans[150000];
Double p,r,t;
void Dfs (int u,int depth)
{
int i;
if (Tree[u].size () ==0)
{
mm=min (mm,depth);
ans[depth]++;
return;
}
For (I=0;i<tree[u].size (); i++)
{
dfs (tree[u][i],depth+1);
}
}
int main ()
{
int i,k,j,num;
cin>>n>>p>>r;
for (i=0;i<n;i++)
{
cin>>k;
for (j=0;j<k;j++)
{
cin>>num;
Tree[i].push_back (num);
}
}
mm=999999;
DFS (0,1);
T=1;
for (i=1;i<mm;i++)
t=t* (1+r*1.0/100);
printf ("%.4lf%d\n", t*p,ans[mm]);
return 0;
}