PAT 1004 Counting Leaves

题目描述

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.

题目大意
  一个家庭层级通常被表示成一棵家谱树。你的任务是去计算没有孩子的家庭成员的数目。

输入说明
  每一个输入文件包含一个测试用例。每个测试用例的第一行含有0<N<100,它是树中节点的数目，M（<N），它是非叶子节点的数目。接下来的M行，每一行以如下形式出现：

ID K ID[1] ID[2] ... ID[K]

  ID是一个代表非叶子节点的两位数，K是它的孩子节点的数目，接下来是一串它所有孩子的两位数ID号。为了简单起见，让我们假设根节点的ID是01。
  输入以N为0结尾，不得处理这种情况。

输出说明
  对于每个测试用例，你应该计算从根节点开始，每一层里没有孩子的家庭成员数目。所有的数字必须输出到一行中，数字之间用空格分隔，并且在行尾没有多余的空格。
  样例表示一棵有两个节点的树，01是根节点，02是它的孩子节点。因此在根节点01这一层上，有0个叶子节点。在下一层，有1个叶子节点。然后我们应该在一行中输出0 1。

题目分析
  这道题考察的知识点是树的遍历。定义一个数组int leaves[105]，其中leaves[i]表示第i层的叶子节点数目。在进行树的遍历时，每遍历到一个节点，先判断它的孩子节点的数目是否为0，若是，说明这个节点是叶子节点，那么就将这个节点的深度所对应的叶子节点数加1。下面提供DFS和BFS两种写法的参考代码。

DFS

#include <bits/stdc++.h>
using namespace std;
int leaves[105] = {};
int maxLevel = 0;
vector<int> tree[105];
void DFS(int s, int level) {
    if (level > maxLevel)
        maxLevel = level;
    if (tree[s].size() == 0) {
        leaves[level]++;
        return;
    }
    for (auto i : tree[s])
        DFS(i, level + 1);
}
int main()
{
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
        int id, k, child;
        cin >> id >> k;
        while (k--) {
            cin >> child;
            tree[id].push_back(child);
        }
    }
    DFS(1, 0);
    for (int i = 0; i < maxLevel; i++)
        cout << leaves[i] << " ";
    cout << leaves[maxLevel] << endl;
    
}

BFS

#include <bits/stdc++.h>
using namespace std;
int leaves[105] = {};
struct Node {
    int level;
    vector<int> child;
}node[105];
int maxLevel = 0;
void BFS(int s) {
    node[s].level = 0;
    queue<int> q;
    q.push(s);
    while (!q.empty()) {
        auto top = q.front();
        q.pop();
        if (node[top].child.size() == 0) {
            maxLevel = node[top].level;
            leaves[node[top].level]++;
            continue;
        }
        for (auto i : node[top].child) {
            node[i].level = node[top].level + 1;
            q.push(i);
        }
    }
}
int main()
{
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
        int id, k, child;
        cin >> id >> k;
        while (k--) {
            cin >> child;
            node[id].child.push_back(child);
        }
    }
    BFS(1);
    for (int i = 0; i < maxLevel; i++)
        cout << leaves[i] << " ";
    cout << leaves[maxLevel] << endl;
    
}