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 01level, 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

题目大意：每个家族都有一个家族关系树，请数出每一辈中无孩子的人数；
解题思路：

• 先用二维vector<int> v(maxn);保存整个家族关系树；
• 利用DFS深度递归，每DFS一次，深度dep+1；
• 边界条件是当前根结点的孩子为空，即v[root].size() == 0条件成立，那么当前dep对应的cnt[dep]++;
• 到达递归边界时，保存最大深度；
• 最后输出cnt数组即可。

  #include <iostream>
  #include <vector>
  using namespace std;
  #define rep(i,j,k) for(int i=j;i<k;i++)
  
  vector<int> v[100];
  int cnt[100]={0},maxDep=-1;
  
  void dfs(int root,int dep){
      if(v[root].size() == 0){
          cnt[dep]++;
          if(maxDep < dep)
              maxDep = dep;
          return;
      }
      rep(i,0,v[root].size()){
          dfs(v[root][i],dep+1);
      }
  }
  int main(){
      int n,m,k,id,c;
      cin>>n>>m;
      rep(i,1,m+1){
          cin>>id>>k;
          rep(j,0,k){
              cin>>c;
              v[id].push_back(c);
          }
      }
      dfs(1,0);
      printf("%d",cnt[0]);
      rep(i,1,maxDep+1)   printf(" %d",cnt[i]);
      return 0;
  }