PAT-甲级-1004

1004. Counting Leaves (30)

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 1

【解析】
第一个给出的数字代表总共的结点数，第二个数字代表的是非叶子结点数。我们要统计的是每一层的叶子结点数有多少，这里可以用dfs和bfs都可以做。dfs的话，就是不断的向下搜索遇到叶子结点的时候该层数叶子结点加1，然后返回。bfs的话也差不多的就是不断的入队和出队。此处说明一下代码来自网络学习。

dfs

#include
#include
#include
using namespace std;
vectora[100];
int b[100],maxdepth=-1;
void dfs(int n,int depth)
{
    if(a[n].size()==0)//判断该结点以下是不是空的
    {
    b[depth]++;//该层数的结点数+1
    maxdepth=max(maxdepth,depth);//求最大的层数有多深
    return;
    }
    for(int i=0;i

bfs

#include
#include
#include
#include
#include
using namespace std;
int level[100],a[100],maxlevel=-1;//level表示所在的层数，maxlevel表示最大的层数
vectorv[100];
void bfs()
{
    queueq;
    q.push(1);//先让根结点入队
    level[1]=0;//根结点所在层数为0
    while(!q.empty())
    {
        int index=q.front();//将队列中的结点的下标分批次的给index
        q.pop();//弹出元素
        maxlevel=max(level[index],maxlevel);
        if(v[index].size()==0)//如果该结点下的元素为空则该层数的空结点数+1
        {
            a[level[index]]++;
        }
        for(int i=0;i