PAT 1004 Counting Leaves（结构体）

标签：mes   leave   follow   not   ber   std   case   specific   math   

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 0leaf 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

题意：输入n个节点，m个非叶子结点
m行，节点id，子节点个数k，k个节点id

 1 #include<iostream>
 2 using namespace std;
 3 
 4 struct Node
 5 {
 6     int father;
 7     int level;
 8     bool leaf;
 9 };
10 Node no[105];
11 int level[105];//记录每层有多少叶子结点
12 int main()
13 {
14     int n,m,id,k,ca,maxle = 1;//如果maxle = -1，测试点2会出错
15     cin >> n >> m;
16     //初始化
17     for(int i = 0;i <= n;i++)
18     {
19         no[i].father = 0;
20         no[i].level = 0;
21         no[i].leaf = 1;
22     }
23     no[1].level = 1;//第一层
24     for(int i = 0;i < m;i++)
25     {
26         cin >> id >> k;
27         no[id].leaf = 0;
28         for(int j = 0;j < k;j++)
29         {
30             cin >> ca;
31             no[ca].father = id;
32         }
33     }
34 
35     for(int i = 1;i <= n;i++)
36     {
37         for(int j = 1;j <= n;j++)
38         {
39             if(no[j].father == i)//父节点
40             {
41                 no[j].level = no[i].level + 1;//层数为父节点的下一层
42                 if(no[j].level > maxle)
43                     maxle = no[j].level;//更新最大层数
44             }
45         }
46         if(no[i].leaf == 1)
47             level[no[i].level]++;//更新每层的叶子结点
48     }
49     for(int i = 1;i < maxle;i++)
50         cout << level[i] << " ";
51     cout << level[maxle] << endl;
52     return 0;
53 }

我一直错一个测试点2，因为把最大层数的初始值定义成了-1，改成1就对了。。。。。

PAT 1004 Counting Leaves（结构体）

标签：mes   leave   follow   not   ber   std   case   specific   math   

原文地址：https://www.cnblogs.com/lu1nacy/p/10046711.html