PAT Advanced 1004 Counting Leaves （30）

题目描述：

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

算法描述：DFS

#include<iostream>
#include<vector>
using namespace std;

vector<int> son[100];
int n, m; // 结点总数  非叶子结点数
int maxlev, eachlev[100] = {0}; // 最大层次   每层叶子结点数量

void dfs(int curid, int curlev)
{
    if(curlev > maxlev) maxlev = curlev; //更新最大层次
    if(son[curid].size() > 0) // 有孩子结点
    {
        for(auto x : son[curid])
        {
            dfs(x, curlev + 1);
        }
    }
    else // 没有孩子结点
    {
        eachlev[curlev] ++;
    }
}

int main()
{
    cin >> n >> m;
    for(int i = 0 ; i < m ; i ++)
    {
        int id, k;
        cin >> id >> k;
        for(int j = 0 ; j < k ; j ++)
        {
            int x;
            cin >> x;
            son[id].push_back(x);
        }
    }
    
    dfs(1, 1);
    
    for(int i = 1 ; i < maxlev ; i ++)
        cout << eachlev[i] << ' ';
    cout << eachlev[maxlev];
    return 0;
}