1079 Total Sales of Supply Chain (25)

Posted mr-stn

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了1079 Total Sales of Supply Chain (25)相关的知识,希望对你有一定的参考价值。

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 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
 
往dfs上面套就行
 1 #include<iostream>
 2 #include<vector>
 3 #include<cmath>
 4 using namespace std;
 5 double total=0.0;
 6 vector<vector<int>> v;
 7 vector<int> w;
 8 double p, r;
 9 void dfs(int root, int depth){
10   if(v[root].size()==0){
11     total += p*pow(1.0+1.0*r/100, depth)*w[root];
12     return;
13   }
14   for(int i=0; i<v[root].size(); i++) dfs(v[root][i], depth+1);
15 }
16 int main(){
17   int n, i;
18   cin>>n>>p>>r;
19   v.resize(n);
20   w.resize(n);
21   for(i=0; i<n; i++){
22     int k, temp;
23     cin>>k;
24     if(k==0){cin>>temp; w[i]=temp;}
25     for(int j=0; j<k; j++){
26       scanf("%d", &temp);
27       v[i].push_back(temp);
28     }
29   }
30   dfs(0, 0);
31   printf("%.1f", total);
32   return 0;
33 }

 

以上是关于1079 Total Sales of Supply Chain (25)的主要内容,如果未能解决你的问题,请参考以下文章

1079. Total Sales of Supply Chain (25)

PAT 1079 Total Sales of Supply Chain[比较]

1079 total sales of supply chain

1079 Total Sales of Supply Chain

PAT (Advanced Level) 1079. Total Sales of Supply Chain (25)

PAT Advanced 1079 Total Sales of Supply Chain (25) [DFS,BFS,树的遍历]