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:
2 101 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
1 #include2 #include 3 #include 4 5 using namespace std; 6 7 //给出一棵树，问每一层的叶子结点数量 8 //使用BFS或者DFS 9 10 vector >nodes(1001);11 vector depth(1001);12 int maxDepth = -1;13 14 void DFS(int index, int h)15 {16 maxDepth = maxDepth > h ? maxDepth : h;17 if (nodes[index].size() == 0)//data[index].size() == 0)//即为叶子结点18 depth[h]++;//层数19 20 for (int i = 0; i < nodes[index].size(); ++i)21 DFS(nodes[index][i], h + 1);22 }23 24 void BFS( )25 {26 queue q;27 q.push(1);28 vector level(1001, 0);//记录节点层数29 while (!q.empty())30 {31 int index = q.front();32 q.pop();33 maxDepth = maxDepth > level[index] ? maxDepth : level[index];//存储最大的层数34 if (nodes[index].size() == 0)//此节点为叶子节点35 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36 for (int i = 0; i < nodes[index].size(); ++i)37 {38 level[nodes[index][i]] = level[index] + 1;//孩子结点层数比父节点多一层39 q.push(nodes[index][i]);//将其孩子全部存入40 }41 }42 }43 44 45 46 int main()47 {48 int N, M;//N为节点数目49 cin >> N >> M; 50 for (int i = 0; i < M; ++i)51 {52 int ID, k, a;53 cin >> ID >> k;54 for (int j = 0; j < k; ++j)55 {56 cin >> a;57 nodes[ID].push_back(a);//即为一个节点底下所挂的子节点58 }59 }60 61 //DFS(1,0);62 BFS( );63 cout << depth[0];64 for (int i = 1; i <= maxDepth; ++i)65 cout << " " << depth[i];66 cout << endl;67 68 return 0;69 70 }