1004. Counting Leaves（30）—PAT 甲级

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

题目大意：计算树的每一层上有多少个叶子节点并按层输出
分析：使用深度有限搜索（DFS）递归遍历树上每一个节点的孩子节点，如果这个节点没有孩子节点，就逐层返回。child[i]集合记录每个节点的孩子节点，leaf[i]数组记录每一层上的叶子节点，max_h记录最大层数，层数从1开始。也可以使用广度优先搜索（BFS）,不同之处是DFS使用集合，BFS使用队列，且BFS不用使用递归，只需要第一个节点入队列后判断队列非空即可。

//DFS求叶子节点
#include <iostream>
#include <vector>
using namespace std;
int leaf[100],max_h=1;
vector<int> child[100];
void DFS(int id_num,int h)
{
    if(max_h<h) max_h=h;
    int k=child[id_num].size();
    if(k==0){
        leaf[h]+=1;
        return;
    }
    for(int i=0;i<k;i++)
    {
        DFS(child[id_num][i],h+1);
    }
}
int main() {
    int n,m;
    scanf("%d %d",&n,&m);
    int id_num,k,id;
    for(int i=0;i<m;i++){
        scanf("%d %d",&id_num,&k);
        for(int j=0;j<k;j++){
            scanf("%d",&id);
            child[id_num].push_back(id);
        }
    }
    DFS(1,1);
    for(int i=1;i<=max_h;i++){
        if(i!=1) printf(" ");
        printf("%d",leaf[i]);
    }
    printf("\n");
    return 0;
}