PAT_A1004#Counting Leaves

时间：2019-06-28 17:11:51 阅读：125 评论：0 收藏：0 [点我收藏+]

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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 $0, the number of nodes in a tree, and$
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

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:

 1 /*
 2 time: 2019-06-28 16:47:33
 3 problem: PAT_A1004#Counting Leaves
 4 AC: 16:22
 5 
 6 题目大意：
 7 统计各层的叶子结点个数
 8 输入：
 9 第一行给出，结点数N<100，分支结点数M
10 接下来M行，结点id，孩子数k，孩子id（1~n，root==1）
11 多组输入样例
12 
13 基本思路：
14 遍历并记录各层叶子结点数即可
15 */
16 #include<cstdio>
17 #include<vector>
18 using namespace std;
19 const int M=110;
20 vector<int> tree[M];
21 int leaf[M],level=0;
22 
23 void Travel(int root, int hight)
24 {
25     if(tree[root].size()==0)
26     {
27         if(hight > level)
28             level = hight;
29         leaf[hight]++;
30         return;
31     }
32     for(int i=0; i<tree[root].size(); i++)
33         Travel(tree[root][i],hight+1);
34 }
35 
36 int main()
37 {
38 #ifdef ONLINE_JUDGE
39 #else
40     freopen("Test.txt", "r", stdin);
41 #endif // ONLINE_JUDGE
42 
43     int n,m;
44     while(~scanf("%d", &n))
45     {
46         level=0;
47         for(int i=1; i<=n; i++){
48             tree[i].clear();
49             leaf[i]=0;
50         }
51         scanf("%d", &m);
52         for(int i=0; i<m; i++)
53         {
54             int id,k,kid;
55             scanf("%d%d", &id,&k);
56             for(int j=0; j<k; j++)
57             {
58                 scanf("%d", &kid);
59                 tree[id].push_back(kid);
60             }
61         }
62         Travel(1,0);
63         for(int i=0; i<=level; i++)
64             printf("%d%c", leaf[i], i==level?‘\n‘:‘ ‘);
65 
66     }
67 
68     return 0;
69 }

PAT_A1004#Counting Leaves

原文：https://www.cnblogs.com/blue-lin/p/11103632.html

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