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A supply chain is a network of retailers（零售商）, distributors（經銷商）, and suppliers（供應商）-- 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 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?(≤10?5??), 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:

K?i???ID[1] ID[2] ... ID[K?i??]

where in the?i-th line,?K?i???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.?K?j???being 0 means that the?j-th member is a retailer, then instead the total amount of the product will be given after?K?j??. 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?10?10??.

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

總結：

這是一道典型的樹的遍歷的圖，與1053類似，沒有遇到大的麻煩，代碼格式DFS：設置邊界——樹的遞歸

鏈接：【2019暑假刷題筆記-樹的遍歷】1053 Path of Equal Weight

代碼：?

#include <stdio.h> #include <vector> #include <math.h> using namespace std; const int maxn=100010; int total,count,counter; double price,r,sum=0; struct Node{int data;vector<int> child; }node[maxn];void DFS(int index,int depth){if(node[index].child.size()==0){sum += node[index].data*pow(1+r,depth); return;}for(int i=0;i<node[index].child.size();i++){ //這一步也很重要深度搜索的命脈 int child=node[index].child[i];DFS(child,depth+1); }} int main(){scanf("%d %lf %lf",&total,&price,&r);r /= 100;int v,child,data;count=0;for(int i=0;i<total;i++){scanf("%d",&v);if(v==0){scanf("%d",&data);node[i].data=data;count++;}for(int j=0;j<v;j++){scanf("%d",&child);node[i].child.push_back(child);}}DFS(0,0);//傳入的參數有結點編號，深度printf("%.1lf\n",price*sum);return 0; }

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總結

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