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序：

又是一道30分的题，仔细一看，这是一道层次遍历多叉树的问题，用queue应该可以搞定。

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

Input Specification:

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.

The input ends with N being 0. That case must NOT be processed.

Output Specification:

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

题目概述：

给出一棵树的节点数，以及所有爸爸节点及其的儿子节点。求每一层中的叶子节点。

分析：

1.找根，根节点是唯一一个不在儿子节点的节点，用一个数组进行0，1记录 2.存储，这里我用了一个map，把每个爸爸和对应的儿子序列对应 3.遍历，用queue遍历，一层层遍历。若遇到节点无儿子，cnt++；否则，将其儿子push进queue，等待下一层遍历

代码如下：

//层次遍历&队列//找根#include
   
    using namespace std;int visit[110];unordered_map
    
     
      > mp;int main(){
       int n, m;    scanf("%d%d", &n, &m);    int father, k, temp;    for(int i = 0; i < m; i++)    {
           scanf("%d%d", &father, &k);        for(int j = 0; j < k; j++)        {
               scanf("%d", &temp);            visit[temp] = 1;            mp[father].push_back(temp);        }    }        int root = 1;    while(visit[root] != 0) root++;        queue
      
        qq;    qq.push(root);            int floor = 1;    while(!qq.empty())    {
           int len = qq.size();        int cnt = 0, ft;        for(int i = 0; i < len; i++)        {
               ft = qq.front();            qq.pop();            if(mp[ft].size() == 0) cnt++;            else            {
                   for(int j = 0; j < mp[ft].size(); j++)                    qq.push(mp[ft][j]);            }        }                if(floor == 1) printf("%d", cnt);        else printf(" %d", cnt);                floor++;    }    printf("\n");    return 0;}
      
     
    
   

总结：

1.queue中的front()是取队首元素，pop()踢掉队首元素 2.用floor作为表示，区分输出的空格 3.root应该从1开始遍历，因为题目中的节点序号对应 4.题目中说01固定为根，但我们给出的是通解。我们的解法可以适用不定根的情况。

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