A1004 Counting Leaves (30 point(s))

每层叶结点数

1. 原文

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

2. 解析

从叶节点开始dfs遍历，遇到叶节点时，更新当前叶所在层的叶节点数 level[deep]++;

3. AC代码

#include<cstdio>
#include<vector>
using namespace std;
vector<int> v[110];
vector<int> level[110];
int maxdeep=-1;
void dfs(int root,int deep)
{
	if (deep>maxdeep)
	{
		maxdeep=deep;
	}
	if (v[root].size()==0)
	{
		level[deep].push_back(root);
		return;
	}
	for (int i = 0; i < v[root].size(); ++i)
	{
		dfs(v[root][i],deep+1);
	}
}
int main()
{
	int n,m,k,id,kid;
	scanf("%d%d",&n,&m);
	while(m--)
	{
		scanf("%d%d",&id,&k);
		for (int i = 0; i < k; ++i)
		{
			scanf("%d",&kid);
			v[id].push_back(kid);
		}
	}
	dfs(1,0);
	int key=0;
	for (int i = 0; i <= maxdeep; ++i)
	{
		if (key>0)
		{
			printf(" ");
		}else{
			key++;
		}
		printf("%d",(int)level[i].size());
	}
	printf("\n");
	return 0;
}