【PAT甲级 - C++题解】1004 Counting Leaves

1004 Counting Leaves

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
• 1
• 2

Sample Output:

0 1

题意

第一行给定一个 N NN 和 M MM ，分别表示该树的结点总数量以及非叶子结点数量。

接下来的 M MM 行，输入每个非叶子结点的孩子结点，格式为：

ID K ID[1] ID[2] ... ID[K]ID 表示非叶子结点的编号，K KK 表示该结点有 K KK 个孩子结点，然后紧跟 K KK 个孩子结点的编号。

让我们统计该树中每一层的叶子结点个数，并输出。

如下图所示，cnt 表示每一层的叶子结点数：

思路

1. 用链式前向星的方法存储该树。
2. 递归更新每层叶子结点的数量，因为题目给定根结点是 1 ，所以从 1 开始向下递归，去递归它的孩子结点。如果在递归过程中 h[u] == -1 即结点 u 没有孩子结点，则将该层的叶子节点数加 1 。
3. 最后输出每层的叶子结点数量。

代码

#include<bits/stdc++.h>
using namespace std;
const int N = 110;
int h[N], ne[N], e[N], idx;    //邻接表
int cnt[N];     //计算每一层的叶子结点数
int max_depth;  //树的最大深度
//链式前向星，往a向b加一条边
void add(int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}
//递归更新每层的叶子结点数量
void dfs(int u, int depth)
{
    //如果当前结点已经是叶子结点，则进行更新操作
    if (h[u] == -1)
    {
        max_depth = max(max_depth, depth);
        cnt[depth]++;
        return;
    }
    //递归遍历其孩子结点
    for (int i = h[u]; ~i; i = ne[i])
        dfs(e[i], depth + 1);
}
int main()
{
    int n, m;
    cin >> n >> m;
    //输入结点之间的关系
    memset(h, -1, sizeof h);
    while (m--)
    {
        int a, b, k;
        cin >> a >> k;
        while (k--)
        {
            cin >> b;
            add(a, b);
        }
    }
    dfs(1, 0);  //根结点为1
    //输出每层的叶子结点数
    cout << cnt[0];
    for (int i = 1; i <= max_depth; i++)
        cout << " " << cnt[i];
    cout << endl;
    return 0;
}