1004 Counting Leaves

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 $0, the number of nodes in a tree, and$

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

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

//input :树中总结点数N 非叶子结点数M
//       非叶子结点值ID 此非叶子结点的孩子数k 此非叶子结点的孩子值id[i]  （M个非叶子结点）
//output: 每层的叶子结点数量
//解题关键:确定每个结点有无孩子结点的标志值tag，每个结点的高度，设置父节点参数nodes[child].parent = ID;
//问题：存在非叶子节点的hight在读入当前行时还未知的情况，可以先存入结点的父子关系，然后输入结束再计算hight
/*问题例子：9 4

            01 2 02 03

            04 3 07 08 09

            03 2 05 06

            02 1 04
*/

#include<stdio.h>
#include<iostream>

using namespace std;

typedef struct Node{
    int hight;//结点所处高度
    int tag;// tag为0，则没有孩子，为1则有孩子
    int parent;//父节点
};

Node nodes[100]; //初始化是100个

int main()
{
    /*step1:构建树*/

    int N; //树中总结点数
    int M; //树中非叶子结点总数
    cin >> N;
    cin >> M;
    //初始化
    for(int i = 1;i <= N;i++)
    {
        nodes[i].hight = 1;
        nodes[i].tag = 0;
        nodes[i].parent = 0;
    }
    //存入父子结点关系
    int ID,k,child;
    for(int i = 0;i < M;i++)
    {
        cin >> ID;
        cin >> k;
        nodes[ID].tag = 1;
        for(int j = 0;j < k;j++)
        {
            cin >> child;
            nodes[child].parent = ID;
            //nodes[child].hight =  nodes[ID].hight + 1;
        }
    }
    //存入高度
    for(int i = 1;i <=N;i++)
    {
        for(int j = i + 1;j <= N;j++)
        {
            if(nodes[j].parent == i)
            {
                nodes[j].hight = nodes[i].hight + 1;
            }
        }
    }

    /*step2:统计每层的叶子结点数，并输出*/
    //确定树的高度
    int maxh = 1;
    for(int i = 1;i <= N;i++)
    {
        if (maxh < nodes[i].hight)
        {
            maxh = nodes[i].hight;
        }
    }
    //确定每层高度的叶子节点数
    int result[100] = {0};//每层高度的叶子节点数
    for(int i = 1;i <= N;i++)
    {
        if (nodes[i].tag == 0)
        {
            result[nodes[i].hight]++;
        }
    }
    //输出
    for(int i = 1;i < maxh;i++)
    {
        cout << result[i] << " ";
    }
    cout << result[maxh]<<endl;

    return 0;
}