1090. Highest Price in Supply Chain (25)

#include<iostream>  
#include<queue>  
#include<vector>  
#include<algorithm>  
using namespace std;  
const int N = 1000010;  
struct node{  
  vector<int> child;  
  int level;  
}node[N];  
void levelorder(int root){  
  queue<int> q;  
  node[root].level = 0;  
  q.push(root);  
  while(!q.empty()){  
    int p = q.front();  
    q.pop();  
    for(int i = 0; i < node[p].child.size(); i++){  
      int temp = node[p].child[i];  
      q.push(temp);  
      node[temp].level = node[p].level + 1;  
    }  
  }  
}  
int main(){  
  int n ;  
  double p , r;  
  int root;  
  cin>>n>>p>>r; 
  for(int i = 0; i < n; i++){  
    int temp;  
    scanf("%d",&temp);  
    if(temp == -1) root = i;  
    else{  
      node[temp].child.push_back(i);  
    }  
  }
  if(n == 1){
  	printf("%.2lf %d",p,n);
  	return 0;
  }   
  levelorder(root);  
  int num = 0, maxlevel = 0;  
  double maxprice = p;  
  for(int i = 0; i < N; i++){  
    if(maxlevel < node[i].level) maxlevel = node[i].level;  
  }  
  for(int i = 0; i < N; i++){  
    if(maxlevel == node[i].level) num++; // 当只有一个点的话num = N，会导致一个1分的点不过
  }  
  for(int i = 0; i < maxlevel; i++){  
    maxprice *=(1.0 + r/100);  
  }  
  printf("%.2lf %d",maxprice,num);  
  return 0;  
}  

当只有一个点的话