题目链接：（PAT甲级）1004 Counting Leaves （30 分）
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题解：其实这题就是树的层次遍历，dfs求出结点的层数。一开始，边输入边根据父结点的层数来求该结点的层数，得了19分，这个思路没有考虑到父结点的层数也许此时还没算出来。毕竟，输入时没什么规律。所以我们先得把父子关系存起来，输入结束后再来计算层数。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <cmath>
using namespace std;
const int maxn=110;
vector<int> v[maxn];
int layer[maxn];
int mx=-1;void dfs(int id,int lay){mx=max(mx,lay);if(v[id].size()==0){layer[lay]++;return ; } for(int i=0;i<v[id].size();i++){int x=v[id][i];dfs(x,lay+1);}
}int main(){int n,m;while(~scanf("%d%d",&n,&m)&&n){memset(layer,0,sizeof(layer));for(int i=1;i<=n;i++) v[i].clear();int id,k,a;while(m--){scanf("%d%d",&id,&k);while(k--){scanf("%d",&a);v[id].push_back(a);}}dfs(1,0);printf("%d",layer[0]);for(int i=1;i<=mx;i++){printf(" %d",layer[i]);}printf("\n");}return 0;
}