PAT -- 甲级1004（1004 Counting Leaves）

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 $, the number of nodes in a tree, and (), the number of non-leaf nodes. Then lines follow, each in the format:$

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 $being 0. That case must NOT be processed.$

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

给你n个结点，当中有m个父亲节点，接下来又m行，每行的分别给父亲节点father，其k个儿子数，接下来k个数代表father的儿子节点，求该棵以1为根节点的树每层的叶子节点数。

思路：DFS遍历这颗树，注意m为0的时候相当于有n个根节点，所以特判一下输出N。

Code：

#include <bits/stdc++.h>

using namespace std;
int N,M;
vector<int> Tree[105];
map<int, int > vis;

void DFS (int idx, int k) {
  if (Tree[idx].size() == 0) {
    vis[k]++;
    return ;
  }
  for (int i = 0; i < Tree[idx].size(); i++) {
    DFS(Tree[idx][i], k + 1);
  }
}

int main() {
  int father, son, k;
  cin >> N >>M;
  while (M--) { //build
    cin >> father >> k;
    while (k--) {
      cin >> son;
      Tree[father].push_back(son);
    }
  }
  if (M == 0) {
    cout << N << endl;
  } else {
    DFS(1, 1);
    cout << vis[1];
    for (int i = 2; i <= vis.size(); i++) {
      cout << " " << vis[i];
    }
    cout << endl;
  }
  return 0;
}