1004 Counting Leaves (30 point(s))

本文介绍了一种使用BFS算法和邻接链表存储结构来解决计数家族树中每层叶子节点数量的问题。通过一维数组模拟队列实现DFS，从根节点开始遍历，根据节点是否有孩子进行相应的操作，最终输出每一层的叶子节点数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

题目大意：

给一颗树，求每一层叶子节点数；

设计思路：

BFS

邻接链表（数据较少，直接用了二维数组模拟）存取树
队列（一维数组模拟）实现 DFS，根节点入队
节点出队时，根据有无孩子进队
- 无孩子节点进队，累加当前层的叶子节点数
- 有孩子节点进队，根据出队父亲节点的层数，记录孩子节点的层数
输出每一层的叶子节点数

注：不能在建立树的过程中根据父亲节点记录孩子节点的层数，因为题目未保证输入节点的过程有序（毕竟 30 point(s)，否则太简单），所以不能保证当前的父亲节点已经记录了自己的层数，所以不能在建立树的过程中根据父亲节点记录其孩子节点的层数

编译器：C (gcc)

#include <stdio.h>

int main(void)
{
        int node[100][100] = {0}, depth[100] = {0};
        int n, m, id, cid, k;
        int i, j, count[100] = {0}, queue[100] = {0};

        scanf("%d%d", &n, &m);
        for (i = 0; i < m; i++) {
                scanf("%d%d", &id, &k);
                for (j = 0; j < k; j++) {
                        scanf("%d", &cid);
                        node[id][j] = cid;
                        //depth[cid] = level[id] + 1;
                }
        }

        depth[1] = 1;
        queue[0] = 1;
        int *p = queue, *q = p + 1, maxdepth = -1;
        for (; p < q; p++) {
                for (k = 0; k < 100 && node[*p][k] != 0; k++) {
                                *q = node[*p][k];
                                depth[*q] = depth[*p] + 1;
                                if (depth[*q] > maxdepth)
                                        maxdepth = depth[*q];
                                q++;
                }
                if (k == 0)
                        count[depth[*p]]++;
        }

        printf("%d", count[1]);
        for (i = 2; i <= maxdepth; i++)
                printf(" %d", count[i]);

        return 0;
}