【PAT】甲级1004. Counting Leaves (30)

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.
Input

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.

Output

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

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> v[100];
int book[100], maxdepth = -1;
void dfs(int index, int depth) {
    if(v[index].size() == 0) {
        book[depth]++;
        maxdepth = max(maxdepth, depth);
        return ;
    }
    for(int i = 0; i < v[index].size(); i++)
        dfs(v[index][i], depth + 1);
}
int main() {
    int n, m, k, node, c;
    scanf("%d %d", &n, &m);
    for(int i = 0; i < m; i++) {
        scanf("%d %d",&node, &k);
        for(int j = 0; j < k; j++) {
            scanf("%d", &c);
            v[node].push_back(c);
        }
    }
    dfs(1, 0);
    printf("%d", book[0]);
    for(int i = 1; i <= maxdepth; i++)
        printf(" %d", book[i]);
    return 0;

一种输入的样例：
输入的含义是：
八个结点\ 四个非空结点
本节点编号\ 该节点孩子个数\ 孩子节点编号

输出的含义：
每一层叶子节点的个数

关于这种样例的示意图解释：

感谢柳婼的代码.