PAT甲级——A1004 Counting Leaves

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, the number of nodes in a tree, and M (<), 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 01level, 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

即遍历整颗数，使用DFS或者DFS
用数组记录每个节点的子节点是谁

 #include <iostream>

 #include <vector>

 #include <queue>

 using namespace std;

 //给出一棵树，问每一层的叶子结点数量

 //使用BFS或者DFS

 vector<vector<int>>nodes();

 vector<int>depth();

 int maxDepth = -;

 void DFS(int index, int h)

 {

     maxDepth = maxDepth > h ? maxDepth : h;

     if (nodes[index].size() == )//data[index].size() == 0)//即为叶子结点

         depth[h]++;//层数

     for (int i = ; i < nodes[index].size(); ++i)

         DFS(nodes[index][i], h + );

 }

 void BFS( )

 {

     queue<int>q;

     q.push();

     vector<int>level(, );//记录节点层数

     while (!q.empty())

     {

         int index = q.front();

         q.pop();

         maxDepth = maxDepth > level[index] ? maxDepth : level[index];//存储最大的层数

         if (nodes[index].size() == )//此节点为叶子节点

             depth[level[index]]++;//之所以要记录每个节点的层数，是因为，同一层有多个节点

         for (int i = ; i < nodes[index].size(); ++i)

         {

             level[nodes[index][i]] = level[index] + ;//孩子结点层数比父节点多一层

             q.push(nodes[index][i]);//将其孩子全部存入

         }

     }

 }

 int main()

 {

     int N, M;//N为节点数目

     cin >> N >> M;

     for (int i = ; i < M; ++i)

     {

         int ID, k, a;

         cin >> ID >> k;

         for (int j = ; j < k; ++j)

         {

             cin >> a;

             nodes[ID].push_back(a);//即为一个节点底下所挂的子节点

         }

     }

     //DFS(1,0);

     BFS( );

     cout << depth[];

     for (int i = ; i <= maxDepth; ++i)

         cout << " " << depth[i];

     cout << endl;

     return ;

 }

巴特西

PAT甲级——A1004 Counting Leaves

Input Specification:

Output Specification:

Sample Input:

Sample Output:

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