1004 Counting Leaves (30分)【BFS，DFS，树的层序遍历】

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<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

题目大意:

 第一行输入告诉你拥有的家庭成员数量n以及非叶子结点数m。接下来是m行，每一行首先输入一个非叶子结点的编号，随后是整数k告诉你该结点拥有k个子结点，接着输入该结点的子节点编号。题目要求你输出各层的非叶子结点个数。

解题思路：

很明显，这道题用到的数据结构是树！

考虑到要我们输出各层的非叶子结点数，我们可以用bfs。首先用一个leaf数组来统计各层的叶子结点数 ，我们从01编号的根节点开始递归，如果当前结点的子结点数为0（即不存在以该结点为起始点的边，这里我用了链式前向星建树，因此条件为head[u]==-1）,则当前层的叶子结点数加1;否则遍历以该结点的所有能到的点。

#include<iostream>
using namespace std;

struct Node {
	int to;
	int next;
}edge[110];
int head[110];
int cnt = 0;
void add(int u, int v)
{
	edge[cnt].to = v;
	edge[cnt].next = head[u];
	head[u] = cnt++;
}

int leaf[110] = { 0 };
int maxlevel = 0;

void bfs(int u,int level)
{
	if (head[u] == -1)
	{
		leaf[level]++;
		if (level > maxlevel)
			maxlevel = level;
	}
	else {
		for (int i = head[u]; ~i; i = edge[i].next)
		{
			bfs(edge[i].to, level+1);
		}
	}
}

int main()
{
	int n, m;
	cin >> n >> m;
	fill(head, head + n+1, -1);
	for (int i = 0; i < m; i++)
	{
		int u,t;
		cin >> u >> t;
		while (t--)
		{
			int v;
			cin >> v;
			add(u, v);
		}
	}
	bfs(1,0);
	for (int i = 0; i <= maxlevel; i++)
	{
		cout << leaf[i];
		if (i == maxlevel)
			cout << endl;
		else cout << " ";
	}
	return 0;
}