1004. Counting Leaves (30)

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本文介绍了一种通过构建邻接矩阵并使用广度优先搜索算法来解决计数家族树中每级无子女成员数量的问题。输入包括树的节点数及非叶节点数，输出为每级无子女成员的数量。

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

思路：把树当做图存储在邻接矩阵中；然后广度优先遍历邻接矩阵；

需要注意的是多push了一个0作为分隔层与层的标志

代码如下：

#include <iostream>
#include<queue>

using namespace std;
#define MAX 101
//根节点是从1开始的，所以数组也按找这样去存
int map[MAX][MAX] = {0};
int i;
int visit_isleaf(int aa)
{
	for (int k = 1;k < MAX;k++)
	{
		if (map[aa][k] == 1)
			return 0;
	}
	return 1;
}
void bfs()
{
	queue<int > q;
	q.push(1);
	q.push(0);
	int counter = 0;
	while (!q.empty())
	{
		int temp = q.front();
		q.pop();
		if (temp == 0)
		{
			if (q.empty())
			{
				cout << counter;
				break;
			}
			else
			{
				q.push(0);
				cout << counter<<" ";
				counter = 0;
			}
		}
		else
		{
			counter += visit_isleaf(temp);
			for (i = 1;i < MAX;i++)
			{
				if (map[temp][i] == 1)
				{

					q.push(i);
				}
			}
		}
	}
}

int main()
{
	int n, m;
	cin >> n >> m;
	for ( i = 0;i < m;i++)
	{
		int id, k;
		cin >> id >> k;
		for (int j = 0;j < k;j++)
		{
			int id2;
			cin >> id2;
			map[id][id2] = 1;
		}
	}

	bfs();
	
	return 0;
}