<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
        
        # Build the tree
        graph = defaultdict(list)
        for u, v in edges:
            graph[u].append(v)
            graph[v].append(u)
        
        # Function to find path using BFS
        def find_path(start, end):
            queue = deque([(start, [])])
            visited = set()
            while queue:
                node, path = queue.popleft()
                if node == end:
                    return path + [node]
                visited.add(node)
                for neighbor in graph[node]:
                    if neighbor not in visited:
                        queue.append((neighbor, path + [node]))
            return []

        # Count the frequency of each node in all paths
        freq = [0] * n
        for start, end in trips:
            path = find_path(start, end)
            for node in path:
                freq[node] += 1

        # Function for dynamic programming to minimize price sum
        def dp(node, parent):
            # dp[node][0] means we do not half the price of node
            # dp[node][1] means we half the price of node
            dp0 = freq[node] * price[node]
            dp1 = dp0 // 2

            for neighbor in graph[node]:
                if neighbor == parent:
                    continue
                child_dp0, child_dp1 = dp(neighbor, node)

                dp0 += min(child_dp0, child_dp1)
                dp1 += child_dp0

            return dp0, dp1

        # Start DP from node 0. (arbitrary root)
        root = 0
        dp0, dp1 = dp(root, -1)
        return min(dp0, dp1)

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