1090 Highest Price in Supply Chain(BFS,DFS)

1090 Highest Price in Supply Chain(BFS,DFS)

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. 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 highest price we 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 they are numbered from 0 to N−1); P, the price given by the root supplier; and r, the percentage rate of price increment for each distributor or retailer. Then the next line contains N numbers, each number Si is the index of the supplier for the i-th member. S_root for the root supplier is defined to be −1. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the highest price we can expect from some retailers, accurate up to 2 decimal places, and the number of retailers that sell at the highest price. There must be one space between the two numbers. It is guaranteed that the price will not exceed 10¹⁰.
Sample Input:

9 1.80 1.00
1 5 4 4 -1 4 5 3 6

Sample Output:

1.85 2

题目大意

供应链是由供应商、供销商和零售商组成，供应链中每个成员只有一个货物提供商（即每一个节点只有一个上级节点），需要我们求出树的最大深度和这个最大深度下有多少个叶子节点。

分析

本题考察树的深度优先遍历和广度优先遍历，遍历的过程中将当前层数与最大层数对比，若大于当前层数则更新最大层数MaxLevel，并将符合要求的节点树cnt置为1，若当前层数等于最大层数MaxLevel,则cnt++

#include 
#include 
#include 
using namespace std;
vector > v;
int MaxLevel=-1,cnt=0;
void dfs(int u,int level){
    if(level>MaxLevel){
        cnt=1;
        MaxLevel=level;
    }else if(level==MaxLevel){
        cnt++;
    }
    for(int i=0;i>n>>p>>r;
    v.resize(n);
    int root;
    for(int i=0;i>val;
        if(val==-1) root=i;
        else v[val].push_back(i);
    }
    dfs(root,0);
    printf("%.2lf %d",p*pow(1+r*0.01,MaxLevel),cnt);

}