PAT 1079 Total Sales of Supply Chain

个人学习记录，代码难免不尽人意。
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 jbeing 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 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

#include 
#include
#include
#include
#include
#include
using namespace std;
  
struct node{
	int data;
	vector<int> child;
	double amount;
	int level;
}; 
node* Node[100010];
node* newnode(int data){
	node* root =new node;
	root->data=data;
	root->child.clear();
	root->amount=0;
	root->level=-1;
	return root;
}
void level(int root){
	queue<int> q;
	q.push(root);
	Node[root]->level=0;
	while(!q.empty()){
		int now=q.front();
		q.pop();
		if(!Node[now]->child.empty()){
			for(int i=0;i<Node[now]->child.size();i++){
				int a=Node[now]->child[i];
				q.push(a);
				Node[a]->level=Node[now]->level+1;
			}
		}
	}
}
int main(){
  int n;
  double p,r;

  scanf("%d%lf%lf",&n,&p,&r);
  for(int i=0;i<n;i++){
  	Node[i]=newnode(i);
  }
  for(int i=0;i<n;i++){
  	int type;
  	scanf("%d",&type);
  	if(type==0){
  		scanf("%lf",&Node[i]->amount);
	  }
	else{
		for(int j=0;j<type;j++){
			int a;scanf("%d",&a);
			Node[i]->child.push_back(a);
		}
	}
  }
  level(0);
  queue<node* > q;
  q.push(Node[0]);
  double sum=0;
  while(!q.empty()){
  	node* no=q.front();
  	q.pop();
  	if(!no->child.empty()){
  		for(int i=0;i<no->child.size();i++){
  				q.push(Node[no->child[i]]);
		  }
	  }
	else{

	    double rate=1+(r/100);
		sum+=p*pow(rate,no->level)*no->amount;

	}
  	
  }
  printf("%.1lf\n",sum);
}

这道题还是蛮简单的，就是我看到题目中说数据不超过10¹⁰,觉得肯定会有数据溢出的测试点，本想着看到错误再改代码的，结果直接通过了，也算是轻松吧！
后来查资料发现，double基本不会溢出，基本可以放心。