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;
}