zoukankan      html  css  js  c++  java
  • 1004. Counting Leaves (30)

    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
    

    题意:给出一棵树,问每一层各有多少个叶子结点。

    分析: bfs解决即可

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <queue>
    #include <string>
    using namespace std;
    
    struct Node{
      int level;
      int firstChild;
      int nextSbiling;
      int childNum;
    };
    
    Node node[101];
    void BFS(){
    
      queue<int> level;
      int NodeTobelevel[101];
      memset(NodeTobelevel,0, sizeof(NodeTobelevel));
    
      node[1].level = 0;
      level.push(1);
      int curId, tmpId,curlevel = 0;
      while(!level.empty()){
    
        curId = level.front();
        level.pop();
        int num = node[curId].childNum;
        curlevel = node[curId].level;
    
        if (num==0)
        {
          NodeTobelevel[node[curId].level]++;
          continue;
        }
    
        tmpId = node[curId].firstChild;
            node[tmpId].level = node[curId].level+1;
            level.push(tmpId);
            num--;
    
    
        while(num--){
          tmpId = node[tmpId].nextSbiling;
          node[tmpId].level = node[curId].level+1;
          level.push(tmpId);
        }
      }
      for (int i = 0; i <= curlevel; ++i){
        if (i==0){
          printf("%d", NodeTobelevel[i]);
        }else{
          printf(" %d", NodeTobelevel[i]);
        }
      }
    }
    
    int main(){
      int N,M;
      scanf("%d%d", &N,&M);
      int curId, firstSibling;
    
      memset(node,0, sizeof(node));
    
      //input node
      for (int i = 0; i < M; ++i)
      {
        scanf("%d",&curId);
        scanf("%d",&(node[curId].childNum));
        scanf("%d",&(node[curId].firstChild));
    
        firstSibling = node[curId].firstChild;
        for (int i = 0; i < node[curId].childNum-1; ++i)
        {
          scanf("%d",&(node[firstSibling].nextSbiling));
          firstSibling = node[firstSibling].nextSbiling;
        }
      }
      BFS();
    }
    
  • 相关阅读:
    02---控制移动底座8
    02---控制移动底座7
    02---控制移动底座6
    02---控制移动底座5
    第四章输入/输出(I/O)4.2PCL中I/O模块及类介绍
    第四章输入/输出(I/O)4.1I/O涉及的设备及相关概念简介
    1.6 opencv视频操作基础
    1.5快速上手OpenCV图像处理
    02---控制移动底座4
    函数cvtColor
  • 原文地址:https://www.cnblogs.com/Dyleaf/p/8725810.html
Copyright © 2011-2022 走看看