1079 Total Sales of Supply Chain (25 分)
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:
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 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
The main idea of the topic: give the price p of the root node, and the increment %r for each next layer. The retailer acts as a leaf node and gives the retailer's sales volume, and other nodes give their sub-nodes. Find the sum of the prices of the goods
Error code:
At first I thought of using a vector to store the sub-node information of each node, and using two vectors to store the id and the number of goods representing the seller's node, and the result was running timeout
#include<bits/stdc++.h>
using namespace std;
double sum = 0;
vector<vector<int>> nextNode;
vector<int> retailer,retailerNum;
double p, r;
int nodeNum;
void DFS(int id,int layer){
for (int i = 0; i < retailer.size();i++){
//和代表销售商的结点匹配
if(id==retailer[i]){
sum += p * pow(0.01 * r + 1, layer) * retailerNum[i];
return;
}
}
for (int i = 0; i < nextNode[id].size(); i++)
{
DFS(nextNode[id][i], layer + 1);//如果有子结点，遍历子结点
}
}
int main(){
cin >> nodeNum >> p >> r;
nextNode.resize(nodeNum + 1);
int a, b;
for (int i = 0; i < nodeNum;i++){
cin >> a;
if(a==0){
cin >> b;
retailer.push_back(i);//存销售商结点的id
retailerNum.push_back(b);//存销售货物
}
for (int j = 0; j < a;j++){
cin >> b;
nextNode[i].push_back(b);//存子结点信息
}
}
DFS(0, 0);
printf("%.1lf", sum);
return 0;
}
After referring to Liushen's code, use a structure to store the id of each node's child node. If there is no child node, store the quantity of goods. When performing DFS, if the node has child nodes, continue to traverse the child nodes, if not, calculate the sum as the recursive exit
AC code
#include<bits/stdc++.h>
using namespace std;
double sum = 0;
struct node{
int num;
vector<int> nextNode;
};
vector<node> v;
double p, r;
int nodeNum;
void DFS(int id,int layer){
if(v[id].nextNode.size()==0){
//没有孩子结点就计算sum，作为递归的出口
sum += p * pow(0.01 * r + 1, layer) * v[id].num;
}
else{
//有孩子结点，则继续遍历
for (int i = 0; i < v[id].nextNode.size(); i++){
DFS(v[id].nextNode[i], layer + 1);
}
}
}
int main(){
cin >> nodeNum >> p >> r;
v.resize(nodeNum + 1);//别少了这一步，会段错误，这句话的细节我目前还没完全弄懂
int a, b;
for (int i = 0; i < nodeNum;i++){
cin >> a;
if(a==0){
cin >> b;
v[i].num = b;//存货物的数量
}
for (int j = 0; j < a;j++){
cin >> b;
v[i].nextNode.push_back(b);//存子结点的信息
}
}
DFS(0, 0);
printf("%.1lf", sum);
return 0;
}
Summary: It is still necessary to accumulate experience in code structure design and data structure selection. According to the second method, one layer of loop in DFS can be removed, which greatly reduces the time complexity