PAT_A1004#Counting Leaves

Source:

PAT A1004 Counting Leaves (30 分)

Description:

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 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

Keys:

• 二叉树的建立和遍历

Code:

 /*
 time: 2019-06-28 16:47:33
 problem: PAT_A1004#Counting Leaves
 AC: 16:22

 题目大意：
 统计各层的叶子结点个数
 输入：
 第一行给出，结点数N<100，分支结点数M
 接下来M行，结点id，孩子数k，孩子id（1~n，root==1）
 多组输入样例

 基本思路：
 遍历并记录各层叶子结点数即可
 */
 #include<cstdio>
 #include<vector>
 using namespace std;
 const int M=;
 vector<int> tree[M];
 int leaf[M],level=;

 void Travel(int root, int hight)
 {
     if(tree[root].size()==)
     {
         if(hight > level)
             level = hight;
         leaf[hight]++;
         return;
     }
     for(int i=; i<tree[root].size(); i++)
         Travel(tree[root][i],hight+);
 }

 int main()
 {
 #ifdef ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif // ONLINE_JUDGE

     int n,m;
     while(~scanf("%d", &n))
     {
         level=;
         for(int i=; i<=n; i++){
             tree[i].clear();
             leaf[i]=;
         }
         scanf("%d", &m);
         for(int i=; i<m; i++)
         {
             int id,k,kid;
             scanf("%d%d", &id,&k);
             for(int j=; j<k; j++)
             {
                 scanf("%d", &kid);
                 tree[id].push_back(kid);
             }
         }
         Travel(,);
         for(int i=; i<=level; i++)
             printf("%d%c", leaf[i], i==level?'\n':' ');

     }

     return ;
 }