PAT 1079 Total Sales of Supply Chain [DFS] [树的遍历]

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:

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

------------------------------------这是题目和解题的分割线------------------------------------

PAT 1090 Highest Price in Supply Chain 和这道题有一点点的相似，都是求层数。不同的是输入方式，以及前者求最大层数，后者多了数据并要求所有叶子结点的层数。

#include
#include
#include

using namespace std;

struct node
{
	vector child;
	int data; //给叶子结点存数据 
}tree[100010];

int n;
double money,rate,sum = 0;

void dfs(int index,int layer)
{
	if(tree[index].child.size()==0) //遍历到了叶子结点 
	{
		int da = tree[index].data;
		sum += pow(1+rate,layer)*da; //每一个叶子结点的零售额什么的相加 
		return;
	}
	for(int i=0;i

也可以不求层数

#include
#include

using namespace std;

vector tree[100010],print;
double start,percent,sum = 0;
int n;

void dfs(int index,double now)
{
	if(tree[index][0]<0)
	{
		//printf("%d\n",tree[index][0]);
		sum += now*(-tree[index][0]);
		return;
	}	
	if(index>n) return;
	for(int i=0;i