PTA 1004 Counting Leaves

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, the number of nodes in a tree, and M (<), 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

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这题没什么难的，可能存父节点不存子节点难想一点，但事实上存子节点也不难，总之随便水水就好。

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代码：

 #include <iostream>
 #include <string>
 using namespace std;
 
 struct NODE{
     int father,depth;
     bool nochild;
 }; 
 
 int main(){
     int n,m,father_id,son_id,son_number,res[]={},depth_max;
     NODE tree[];
     for (int i=;i<=;i++){
         tree[i].father=tree[i].depth=;
         tree[i].nochild=true;
     }
     cin >> n >> m;
     for (int i=;i<m;i++){
         cin >> father_id >> son_number;
         if (son_number!=) tree[father_id].nochild=false;
         for (int j=;j<son_number;j++){
             cin >> son_id;
             tree[son_id].father=father_id;
         }
     }
     for (int i=;i<=n;i++){
         int now=i,level=;
         for (;tree[now].father!=;){
             now=tree[now].father;
             level++;
         }
         tree[i].depth=level;
         depth_max=(depth_max>level)?depth_max:level;
     }
     for (int i=;i<=n;i++){
         if (tree[i].nochild) res[tree[i].depth]++;
     }
     for (int i=;i<depth_max;i++)
         cout << res[i] << " ";
     cout << res[depth_max];
     return ;
 }