1004 Counting Leaves (30分)
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时间限制: 400 ms
内存限制: 64 MB
代码长度限制: 16 KB
思路
使用邻接表存储孩子节点，使用BFS层序遍历判断是否叶子节点，如果该节点邻接表为空，说明是该节点是叶子节点；否则，将其所有孩子节点入队。
1 #include <iostream>
2 #include <stdio.h>
3 #include <vector>
4 #include <queue>
5 #include <string.h>
6
7 using namespace std;
8
9 struct Node
10 {
11 int val;
12 int level = 0;
13 };
14
15 int cnt[110]; //cnt[i]表示第i层的叶子节点数
16 int totalLevel;
17
18 int main()
19 {
20 int n, m, k;
21 while(scanf("%d", &n) != EOF)
22 {
23 if(n == 0)
24 break;
25
26 vector<Node> adjList[110]; //邻接表
27 queue<Node> Q;
28 memset(cnt, 0, sizeof(cnt));
29
30 scanf("%d", &m);
31
32 if(n == 1)
33 {
34 //只有1个节点，直接输出1
35 cout << 1;
36 continue;
37 }
38 for(int i = 0; i < m; ++i)
39 {
40 Node id;
41 int k;
42 scanf("%d%d", &id.val, &k);
43 if(id.val == 01)
44 Q.push(id); //根节点入队
45 for(int i = 1; i <= k; ++i)
46 {
47 Node child;
48 scanf("%d", &child.val);
49 //建立邻接表
50 adjList[id.val].push_back(child);
51 }
52 }
53
54 //层序遍历，BFS
55 while(!Q.empty())
56 {
57 Node p = Q.front();
58 Q.pop();
59
60 //存储最后的总层数，方便最后输出结果
61 totalLevel = p.level;
62
63 if(adjList[p.val].empty())
64 {
65 cnt[p.level]++; //当前层的叶子节点数加一
66 }
67 else
68 {
69 //遍历孩子节点
70 for(int i = 0; i < adjList[p.val].size(); ++i)
71 {
72 adjList[p.val][i].level = p.level + 1;
73 Q.push(adjList[p.val][i]);
74 }
75 }
76 }
77
78 for(int i = 0; i <= totalLevel; ++i)
79 {
80 if(i == totalLevel)
81 cout << cnt[i];
82 else
83 cout << cnt[i] << ' ';
84 }
85
86 }
87
88 return 0;
89 }
