[PAT甲级]1004 Counting Leaves （30 分)

题目

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

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

分析

这道题主要考察的是dfs的运用，因为在树中是不会有回路的，所以dfs过程中不需要辅助数组vis，在dfs中需要考虑level也就是每一层树中叶子节点的个数，所以多加一个level参数。退出条件是遍历到叶子结点（叶子节点没有子树了），同时要存储一个最大的level数，用于输出level数组。
解法中使用二维数组来存储树的结构

代码

#include
#include
using namespace std;
#define maxn 101
vectornode[maxn];
int level[maxn]={0};
int maxlevel=0;

void dfs(int n,int l){
    if(node[n].size()==0){
        level[l]++;
        if(l>maxlevel)
            maxlevel=l;
        return;
    }
    for(int i=0;i>n>>m;
    for(int i=0;i>u>>k;
        for(int j=0;j>x;
            node[u].push_back(x);
        }
    }
    dfs(1,0);
    cout<