PAT 1004 Counting Leaves [DFS] [树的遍历] [每层的叶子结点个数]

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 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

-----------------------------------这是题目和解题的分割线-----------------------------------

求每一层的叶子结点数。

#include<cstdio>
#include<vector>
#include<algorithm>

using namespace std;

vector<int> tree[110],tmp; 
int node[110] = {},sum; 

void dfs(int index,int layer)
{
	sum = max(layer,sum); //记录整棵树的层数，必须取最大值，因为每棵子树的层数不同 
	if(tree[index].size()==0) node[layer]++; //更新每一层的叶子结点数 
	for(int i=0;i<tree[index].size();i++)
		dfs(tree[index][i],layer+1);
}

int main()
{
	int n,m,id,child,x,i,j;
	scanf("%d%d",&n,&m);
	for(i=0;i<m;i++)
	{
		scanf("%d%d",&id,&x);
		for(j=0;j<x;j++)
		{
			scanf("%d",&child);
			tree[id].push_back(child);
		}	
	}
	dfs(1,1); //根结点是1，第一层的编号是1 
	for(i=1;i<=sum;i++)
	{
		printf("%d",node[i]);
		if(i!=sum) printf(" "); //末尾无空格 
	}		
	return 0;
}