1004. Counting Leaves (30)

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本文探讨如何通过家庭谱系树计算无子女的家庭成员数量，提供了一种层次遍历算法来解决实际问题。

1004. Counting Leaves (30)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.

Input

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.

Output

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 l

#include <iostream>
#include <vector>
#include <queue>
using namespace std;
#define MaxNode 105
typedef struct node
{
    int value;
    struct node* Next;
}Node,*pNode;

int main()
{
    
    int N,M;
    cin>>N>>M;
    int ID,K,temp;
    pNode ptemp,temp1;
    Node Son[MaxNode];
    for (int i=1; i<=N; ++i)
    {
        Son[i].Next=NULL;
    }
    for (int i=0; i<M; ++i)
    {
        cin>>ID>>K;
        temp1=&Son[ID];
        for (int k=0; k<K; ++k)
        {
            cin>>temp;
            ptemp=(pNode)new Node;
            ptemp->Next=NULL;
            ptemp->value=temp;
            temp1->Next=ptemp;
            temp1=ptemp;
        }
    }
    
    //层次遍历树
    queue<int> Q;
    int flag=0;
    int num_per_scale=1;        //当前层的结点个数,第一层至少有一个根结点
    int num_next_scale=0;       //下一层结点的个数
    int num_already=0;          //当前层已遍历结点个数
    int temp_value=0;           //当前进行判断的结点编号
    int no_leaf_num=0;              //当前层没有叶子的结点个数
    pNode p;                    //用来遍历每个结点的孩子
    Q.push(1);
    
    while (Q.size()>0)          //还未遍历完所有结点
    {
        temp_value=Q.front();
        Q.pop();
        
        p=Son[temp_value].Next;
        if (p==NULL)
        {
            no_leaf_num++;
        }else
        {
            while (p!=NULL)
            {
                num_next_scale++;
                Q.push((p->value));
                p=p->Next;
            }
        }
        num_already++;
        
        if (num_already==num_per_scale)
        {
            if (flag==0)
            {
                cout<<no_leaf_num;
                flag=1;
            }else
            {
                cout<<" "<<no_leaf_num;
            }
            num_per_scale=num_next_scale;
            num_next_scale=0;
            no_leaf_num=0;
            num_already=0;
        }
    }
    return 0;
}