PAT 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<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 0leaf 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

题意：输入n个节点，m个非叶子结点
m行，节点id，子节点个数k，k个节点id

思路：
定义一个结构体保存其父节点，是否是叶子结点，所在的层数；
感觉这样比dfs简单一点。

#include<iostream>
using namespace std;

struct Node
{
    int father;
    int level;
    bool leaf;
};
Node no[105];
int level[105];//记录每层有多少叶子结点
int main()
{
    int n,m,id,k,ca,maxle = 1;//如果maxle = -1，测试点2会出错
    cin >> n >> m;
    //初始化
    for(int i = 0;i <= n;i++)
    {
        no[i].father = 0;
        no[i].level = 0;
        no[i].leaf = 1;
    }
    no[1].level = 1;//第一层
    for(int i = 0;i < m;i++)
    {
        cin >> id >> k;
        no[id].leaf = 0;
        for(int j = 0;j < k;j++)
        {
            cin >> ca;
            no[ca].father = id;
        }
    }

    for(int i = 1;i <= n;i++)
    {
        for(int j = 1;j <= n;j++)
        {
            if(no[j].father == i)//父节点
            {
                no[j].level = no[i].level + 1;//层数为父节点的下一层
                if(no[j].level > maxle)
                    maxle = no[j].level;//更新最大层数
            }
        }
        if(no[i].leaf == 1)
            level[no[i].level]++;//更新每层的叶子结点
    }
    for(int i = 1;i < maxle;i++)
        cout << level[i] << " ";
    cout << level[maxle] << endl;
    return 0;
}

我一直错一个测试点2，因为把最大层数的初始值定义成了-1，改成1就对了。。。。。