Pat grade a a1079 total sales of supply chain

A supply chain is a network of retailers（ Retailer ）, distributors（ Distributor ）, and suppliers（ supplier ）-- 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 N−1, 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

The question ：

There are suppliers （ The root node ）, Distributor （ Child nodes of non leaf leaves ） And retailers （ leaf node ）, A tree , Only retailers have the quantity of goods （ The weight ）, Give the number of nodes, the price of goods and the percentage that will rise every time you pass through a store , Find the total price of all goods of all retailers of this tree （ The layers of retailers may be different ）.

Ideas ：

and 1090 I think about it , The leaf node has more weights , Build first , Then traverse the tree , Just calculate the maximum value .

Code ：
 

// The tree of the relationship between number and node , Use the static method of tree ; 
#include<iostream>
#include<vector>
#include<cmath>
using namespace std; 
const int maxn=100010;
int child,n,k;
double p,r,ans=0; //ans Should be double  type ; 
struct node{
	double data;// Cargo volume （ The weight ）, Only leaf nodes have ; 
	vector<int > child;
}Node[maxn];

void DFS(int index,int depth){
	if(Node[index].child.size()==0){
		ans+=Node[index].data*pow(1+r,depth);
		return ;// Recursive boundary ;; 
	}
	for(int i=0;i<Node[index].child.size();i++)
		DFS(Node[index].child[i],depth+1);
}

int main( ){
	scanf("%d %lf %lf",&n,&p,&r);
	r/=100;
	// Build up trees , assignment ： 
	for(int i=0;i<n;i++){
		scanf("%d",&k);// Input i Number of sub nodes of node ;
		
		if(k==0){// No child node is a leaf node , Only leaf nodes have cargo volume （ With power ）; 
				scanf("%lf",&Node[i].data);
		}else{
			for(int j=0;j<k;j++){// Child node number ; 
				scanf("%d",&child);
				Node[i].child.push_back(child);
			} 
		}  
	}
	// Traversing the tree ; 
	DFS(0,0);
	printf("%.1f",p*ans);// Price * Floating range = Rising prices , Price * Floating range * Cargo volume = The total value after the increase of total cargo volume , So you can first p Bring up , Calculate the floating amplitude of each node * Cargo volume , Finally, multiply by the original price to get the total value . 
	return 0; 
}