PAT - 甲级 - 1106. Lowest Price in Supply Chain (25)（多叉树）

题目描述：

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 lowest price a customer can expect from some 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 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. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the lowest price we can expect from some retailers, accurate up to 4 decimal places, and the number of retailers that sell at the lowest price. There must be one space between the two numbers. It is guaranteed that the all the prices will not exceed 10¹⁰.

10 1.80 1.00
3 2 3 5
1 9
1 4
1 7
0
2 6 1
1 8
0
0
0

1.8362 2

题目思路：

有1个供应商，n-1个经销商和零售商，每一个零售商到供应商都有唯一的路径，并且没有环，根据题意可以判断这是一棵树。题目要求出从哪个零售商购买商品所需要的花费最少，并且输出这样的零售商的个数。我们只需要从根节点深搜到叶子结点，找出深度最小的叶子结点即可。

如图是题目中给出的数据的图像。

题目代码：

#include <cstdio>
#include <vector>
#include <cmath>
using namespace std;vector<int>v[100000];
int n, a, b, mind = 100000;
int d[100000];
double p, r;
// 搜索每个结点的深度 
void dfs(int root, int de){if(v[root].size() == 0) return ;for(int i = 0; i < v[root].size(); i++){d[v[root][i]] = de+1;dfs(v[root][i], de+1);}
}int main(){scanf("%d%lf%lf",&n, &p, &r);// 数据输入 for(int i = 0; i < n; i++){scanf("%d",&a);for(int j = 0; j < a; j++){scanf("%d",&b);v[i].push_back(b);}}dfs(0,0);// 寻找retailers的最小深度 for(int i = 0; i < n; i++){if(v[i].size() == 0){if(mind > d[i]){mind = d[i];}}}// 记录retailers的深度为最小深度的个数 int ans = 0;for(int i = 0; i < n; i++){if(v[i].size() == 0){if(d[i] == mind){ans ++;}}}// 输出 printf("%.4f ",p*pow((100+r)/100,mind));printf("%d",ans);return 0;
}