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

#include<bits/stdc++.h>

using namespace std;

const int N=110;

vector<int> v[N]; //邻接表存储图
int cnt[N]; //每一深度（层）叶子点数
int maxdepth; //最大深度（层）便于最后输出

void dfs(int u,int depth)
{
    if(!v[u].size())  //没有边，表示叶子结点 
    {
        cnt[depth]++;
        maxdepth=max(maxdepth,depth);
        return; 
    } 
    for(int i=0;i<v[u].size();i++)
        dfs(v[u][i],depth+1);
}

int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    while(m--)
    {
        int id,childnum,child;
        scanf("%d%d",&id,&childnum);
        while(childnum--)
        {
            scanf("%d",&child);
            v[id].push_back(child); 
        }
    }

    dfs(1,0);

    printf("%d",cnt[0]);
    for(int i=1;i<=maxdepth;i++)
        printf(" %d",cnt[i]);
    printf("\n");
    return 0;
 }