PAT_A1079#Total Sales of Supply Chain

Source:

PAT A1079 Total Sales of Supply Chain (25 分)

Description:

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 Pand 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 (≤), 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 1.

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

Keys:

• 二叉树的建立和遍历

Code:

 /*
 time: 2019-03-30 18:34:12
 problem: PAT_A1079#Total Sales of Supply Chain
 AC: 12:50

 题目大意：
 根结点价格为p，各层溢价r%，计算所有叶子结点的价格（单价*数量）
 输入：
 第一行给出：结点数N，p,r
 接下来N行，孩子结点数，孩子编号（0~N-1，root==0）；若孩子数为0，则给出销售数量

 基本思路：
 遍历统计叶子结点的深度并计算价格即可
 */
 #include<cstdio>
 #include<vector>
 #include<cmath>
 using namespace std;
 const int M=1e5+;
 vector<int> tree[M];
 int sell[M]={};
 double total=,p,r;

 void Travel(int root, int hight)
 {
     if(tree[root].size() == )
     {
         total += p*pow((0.01*r+),hight)*sell[root];
         return;
     }
     for(int i=; i<tree[root].size(); i++)
         Travel(tree[root][i],hight+);
 }

 int main()
 {
 #ifdef ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif // ONLINE_JUDGE

     int n,k,kid;
     scanf("%d%lf%lf", &n,&p,&r);
     for(int i=; i<n; i++)
     {
         scanf("%d", &k);
         if(k==)
         {
             scanf("%d", &kid);
             sell[i]=kid;
         }
         for(int j=; j<k; j++)
         {
             scanf("%d", &kid);
             tree[i].push_back(kid);
         }
     }
     Travel(,);
     printf("%.1f", total);

     return ;
 }