PAT 甲级 A1079 Total Sales of Supply Chain

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 (≤105), 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 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

题意：

有供应商（根节点），经销商（非叶子叶子的子节点）和零售商（叶子结点），构成的一棵树，只有零售商有货物量（权重），给出结点的个数和货物价格以及每经过一个店面就要涨的百分比，求这棵树的所有零售商所有货物的总价格（零售商的所在的层可能不一样）。

思路：

和1090思路差不多，叶子结点多了个权重而已，先建树，然后遍历树，算出最大价值就好了。

代码：
 

//编号和结点的关系的树，用树的静态写法; 
#include<iostream>
#include<vector>
#include<cmath>
using namespace std; 
const int maxn=100010;
int child,n,k;
double p,r,ans=0; //ans应该是double 类型; 
struct node{
	double data;//货物量（权重），只有叶子结点有； 
	vector<int > child;
}Node[maxn];

void DFS(int index,int depth){
	if(Node[index].child.size()==0){
		ans+=Node[index].data*pow(1+r,depth);
		return ;//递归边界;; 
	}
	for(int i=0;i<Node[index].child.size();i++)
		DFS(Node[index].child[i],depth+1);
}

int main( ){
	scanf("%d %lf %lf",&n,&p,&r);
	r/=100;
	//建树，赋值： 
	for(int i=0;i<n;i++){
		scanf("%d",&k);//输入i结点下子结点的数目;
		
		if(k==0){//没有子节点就是叶子结点，只有叶子结点有货物量（带权）; 
				scanf("%lf",&Node[i].data);
		}else{
			for(int j=0;j<k;j++){//子节点编号； 
				scanf("%d",&child);
				Node[i].child.push_back(child);
			} 
		}  
	}
	//遍历树; 
	DFS(0,0);
	printf("%.1f",p*ans);//价格*上浮幅度=上涨价格，价格*上浮幅度*货物量=总货物量上涨后的总价值，所以可以先将p提出来，算出各个结点的浮幅度*货物量,最后再乘以原价格也可以得到总价值。 
	return 0; 
}