1004 Counting Leaves（30 分）

1004 Counting Leaves（30 分）

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

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

 

题目意思：

ght跟我说的，看不懂，然后 n，m表示一个有多少个节点，还有m个非叶子节点

然后接下来是m行 每一行都表示这个节点是什么东西，然后有多少个节点

最后输出要求输出每一层的叶子节点个数

代码:

#include
using namespace std;
vectorvec[105],vc;
setst;
int num[105],cnt = 0,maxn = 1;
void DFS(int x,int floor)
{
	if(vec[x].size()==0)
	{
		vc.push_back(x);
		num[floor]++;
		return ;
	}
	for(int j = 0 ; j < vec[x].size(); j++)
	{
		floor++;
		maxn = max(maxn,floor);
		DFS(vec[x][j],floor);
		floor--;
		
	}
	return;
}
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i = 1; i <= m; i++)
	{
		int x,k,y;
		scanf("%d %d",&x,&k);
		for(int j = 1; j <= k; j++)
		{
			scanf("%d",&y);
			{
				vec[x].push_back(y);	
				st.insert(y); 
			}
		}
	}
	int root;
	for(int i =1 ; i <= n; i++)
	{
		if(st.count(i) == 0)
		{
			root = i;
			break;
		}
	}	
	DFS(root,1); 
	for(int i =1; i <= maxn; i++)
	{
		printf("%d%c",num[i]," \n"[i==maxn]);
	} 
	return 0;
}