给定 Q 查询的最小频率的彩色树的子树中不同颜色的计数

题目描述

给定一棵彩色树，每个节点都有一个颜色。定义一个子树的重要性为它包含的不同颜色的数量，定义一组查询为给定一个参数 $Q$，要求找到权值大于等于 $Q$ 的所有节点的子树中，重要性最小的子树，输出该子树中不同颜色的数量。

解题思路

这道题目需要我们能够快速求出一棵树的子树中不同颜色的数量，那么我们可以考虑根据题目特点来寻找解题思路。

首先，我们要注意到这道题目查询的内容是针对子树的。因此，我们可以想到采用树的遍历算法，对于每个节点，都求出它的重要性并保存起来。然后，我们就可以进行查询，找到所有满足条件的节点并对它们的子树进行筛选。

其次，我们注意到题目要求输出重要性最小的子树中不同颜色的数量。这个要求其实比较特殊，我们可以考虑将它们组合起来进行求解。也就是说，我们可以直接按照每个节点的重要性来排序，然后从小到大依次遍历到满足条件的节点即可。在计算节点的重要性时，我们可以使用哈希表来进行存储，这样就能够很快地计算出子树中的不同颜色数量。

代码实现

下面我们给出一份示例代码，以供参考：

class TreeNode:
    def __init__(self, color):
        self.color = color
        self.children = []

def count_colors(node, freq):
    color_counts = {}
    for child in node.children:
        child_counts = count_colors(child, freq)
        for color, count in child_counts.items():
            color_counts[color] = color_counts.get(color, 0) + count
    color_counts[node.color] = color_counts.get(node.color, 0) + 1
    freq_counts[color_counts.get(node.color, 0)] = freq_counts.get(color_counts.get(node.color, 0), 0) + 1
    return color_counts

def find_min_freq_subtree(node, freq, min_freq):
    color_counts = {}
    for child in node.children:
        child_counts = find_min_freq_subtree(child, freq, min_freq)
        for color, count in child_counts.items():
            color_counts[color] = color_counts.get(color, 0) + count
    color_counts[node.color] = color_counts.get(node.color, 0) + 1
    if freq[color_counts.get(node.color, 0)] >= min_freq:
        return color_counts
    else:
        return {}

def get_min_freq_subtree(node, freq, min_freq):
    min_counts = {}
    for child in node.children:
        counts = find_min_freq_subtree(child, freq, min_freq)
        for color, count in counts.items():
            min_counts[color] = min_counts.get(color, 0) + count
    min_counts[node.color] = min_counts.get(node.color, 0) + 1
    return len(min_counts.keys())

if __name__ == '__main__':
    freq = [0] * 1001
    tree = TreeNode(1)
    tree.children.append(TreeNode(2))
    tree.children.append(TreeNode(3))
    tree.children[0].children.append(TreeNode(3))
    tree.children[1].children.append(TreeNode(4))
    tree.children[1].children.append(TreeNode(5))
    tree.children[1].children[0].children.append(TreeNode(6))
    tree.children[1].children[0].children.append(TreeNode(7))
    tree.children[1].children[0].children.append(TreeNode(8))
    tree.children[1].children[0].children[0].children.append(TreeNode(9))
    tree.children[1].children[0].children[0].children.append(TreeNode(10))
    tree.children[1].children[0].children[1].children.append(TreeNode(11))
    tree.children[1].children[0].children[1].children.append(TreeNode(12))
    tree.children[1].children[0].children[1].children.append(TreeNode(13))
    tree.children[1].children[0].children[1].children.append(TreeNode(14))
    tree.children[1].children[1].children.append(TreeNode(6))
    tree.children[1].children[1].children.append(TreeNode(7))
    tree.children[1].children[1].children.append(TreeNode(8))
    tree.children[1].children[1].children.append(TreeNode(15))
    tree.children[1].children[1].children[0].children.append(TreeNode(9))
    tree.children[1].children[1].children[0].children.append(TreeNode(10))
    tree.children[1].children[1].children[3].children.append(TreeNode(16))
    tree.children[1].children[1].children[3].children.append(TreeNode(17))
    tree.children[1].children[1].children[3].children.append(TreeNode(18))
    tree.children[1].children[1].children[3].children.append(TreeNode(19))
    tree.children[1].children[1].children[3].children[0].children.append(TreeNode(20))
    tree.children[1].children[1].children[3].children[0].children.append(TreeNode(21))
    tree.children[1].children[1].children[3].children[0].children.append(TreeNode(22))
    tree.children[1].children[1].children[3].children[0].children.append(TreeNode(23))
    print(get_min_freq_subtree(tree, freq, 5))

结语

彩色树是一种比较特殊的数据结构，在实际问题中并不常见。但是，本题所展示的思路和算法对于解决其他类似问题也是具有借鉴意义的。相信通过本篇文章的学习，大家对于这类题目的解决方法有了更深刻的了解，也能够在编程中得心应手。