<p>There exists an undirected and unrooted tree with <code>n</code> nodes indexed from <code>0</code> to <code>n - 1</code>. You are given the integer <code>n</code> and a 2D integer array <code>edges</code> of length <code>n - 1</code>, where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> in the tree.</p>

<p>Each node has an associated price. You are given an integer array <code>price</code>, where <code>price[i]</code> is the price of the <code>i<sup>th</sup></code> node.</p>

<p>The <strong>price sum</strong> of a given path is the sum of the prices of all nodes lying on that path.</p>

<p>Additionally, you are given a 2D integer array <code>trips</code>, where <code>trips[i] = [start<sub>i</sub>, end<sub>i</sub>]</code> indicates that you start the <code>i<sup>th</sup></code> trip from the node <code>start<sub>i</sub></code> and travel to the node <code>end<sub>i</sub></code> by any path you like.</p>

<p>Before performing your first trip, you can choose some <strong>non-adjacent</strong> nodes and halve the prices.</p>

<p>Return <em>the minimum total price sum to perform all the given trips</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/03/16/diagram2.png" style="width: 541px; height: 181px;" />
<pre>
<strong>Input:</strong> n = 4, edges = [[0,1],[1,2],[1,3]], price = [2,2,10,6], trips = [[0,3],[2,1],[2,3]]
<strong>Output:</strong> 23
<strong>Explanation:</strong> The diagram above denotes the tree after rooting it at node 2. The first part shows the initial tree and the second part shows the tree after choosing nodes 0, 2, and 3, and making their price half.
For the 1<sup>st</sup> trip, we choose path [0,1,3]. The price sum of that path is 1 + 2 + 3 = 6.
For the 2<sup>nd</sup> trip, we choose path [2,1]. The price sum of that path is 2 + 5 = 7.
For the 3<sup>rd</sup> trip, we choose path [2,1,3]. The price sum of that path is 5 + 2 + 3 = 10.
The total price sum of all trips is 6 + 7 + 10 = 23.
It can be proven, that 23 is the minimum answer that we can achieve.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/03/16/diagram3.png" style="width: 456px; height: 111px;" />
<pre>
<strong>Input:</strong> n = 2, edges = [[0,1]], price = [2,2], trips = [[0,0]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The diagram above denotes the tree after rooting it at node 0. The first part shows the initial tree and the second part shows the tree after choosing node 0, and making its price half.
For the 1<sup>st</sup> trip, we choose path [0]. The price sum of that path is 1.
The total price sum of all trips is 1. It can be proven, that 1 is the minimum answer that we can achieve.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 50</code></li>
	<li><code>edges.length == n - 1</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>edges</code> represents a valid tree.</li>
	<li><code>price.length == n</code></li>
	<li><code>price[i]</code> is an even integer.</li>
	<li><code>1 &lt;= price[i] &lt;= 1000</code></li>
	<li><code>1 &lt;= trips.length &lt;= 100</code></li>
	<li><code>0 &lt;= start<sub>i</sub>, end<sub>i</sub>&nbsp;&lt;= n - 1</code></li>
</ul>

class Solution:
    def minimumTotalPrice(self, n: int, edges: List[List[int]], price: List[int], trips: List[List[int]]) -> int:
        from collections import defaultdict, deque
        import math
        
        # Build the graph
        graph = defaultdict(list)
        for u, v in edges:
            graph[u].append(v)
            graph[v].append(u)
        
        # Function to find the path between two nodes using BFS
        def find_path(start, end):
            queue = deque([(start, [start])])
            visited = set()
            while queue:
                node, path = queue.popleft()
                if node == end:
                    return path
                visited.add(node)
                for neighbor in graph[node]:
                    if neighbor not in visited:
                        queue.append((neighbor, path + [neighbor]))
            return []
        
        # Calculate the frequency of each node in all trips
        node_freq = [0] * n
        for start, end in trips:
            path = find_path(start, end)
            for node in path:
                node_freq[node] += 1
        
        # Dynamic programming to find the minimum total price sum
        dp = [[0, 0] for _ in range(n)]
        
        def dfs(node, parent):
            # Calculate the cost when the node is not halved
            dp[node][0] = price[node] * node_freq[node]
            # Calculate the cost when the node is halved
            dp[node][1] = (price[node] // 2) * node_freq[node]
            
            for child in graph[node]:
                if child != parent:
                    dfs(child, node)
                    # Update the cost when the node is not halved
                    dp[node][0] += min(dp[child][0], dp[child][1])
                    # Update the cost when the node is halved
                    dp[node][1] += dp[child][0]
        
        # Start the DFS from an arbitrary root (node 0)
        dfs(0, -1)
        
        # The minimum total price sum is the minimum of the cost when the root is halved and not halved
        return min(dp[0][0], dp[0][1])

# Example usage:
solution = Solution()
n = 4
edges = [[0,1],[1,2],[1,3]]
price = [2,2,10,6]
trips = [[0,3],[2,1],[2,3]]
print(solution.minimumTotalPrice(n, edges, price, trips))  # Output: 23
225