zoukankan      html  css  js  c++  java
  • PAT 1004. Counting Leaves

    PAT 1004. Counting Leaves (30)

    A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.
    Input

    Each input file contains one test case. Each case starts with a line containing 0 < N < 100, the number of nodes in a tree, and M (< N), the number of non-leaf nodes. Then M lines follow, each in the format:

    ID K ID[1] ID[2] ... ID[K]

    where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID's of its children. For the sake of simplicity, let us fix the root ID to be 01.

    Output

    For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.

    The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output "0 1" in a line.

    Sample Input

    2 1
    01 1 02
    

    Sample Output

    0 1
    

    分析

    树的遍历

    代码如下

    #include<iostream>
    #include<vector>
    using namespace std;
    vector<int> result(100,0);
    vector<int> vi[101];
    int maxdepth=-1;
    void dfs(int parent,int depth){
    	if(vi[parent].size()==0){
    		result[depth]++;
    		maxdepth=depth>maxdepth?depth:maxdepth;
    		return;
    	}
    	for(int i=0;i<vi[parent].size();i++)
    	dfs(vi[parent][i],depth+1);
    }
    int main(){
      int n,m,parent,num,child;
      cin>>n>>m;
      for(int i=0;i<m;i++){
      	cin>>parent>>num;
      	for(int j=0;j<num;j++){
      		cin>>child;
      		vi[parent].push_back(child);
    	  }
      }
      dfs(1,0);
      for(int i=0;i<=maxdepth;i++)
      i>0?cout<<" "<<result[i]:cout<<result[i];
      return 0;	
    } 
    
  • 相关阅读:
    asp.net 、C#实现微信企业号OAuth2认证
    node event中 on emit off 的封装
    node
    Express中间件
    旋转的魔方
    通过gulp为requireJs引入的模块添加版本号
    css水平垂直居中(绝对定位居中)
    COLOR 与 COLORREF
    VMware Workstation 安装 vmware tools
    MMIV: starter level 1
  • 原文地址:https://www.cnblogs.com/A-Little-Nut/p/8185431.html
Copyright © 2011-2022 走看看