<p>You are given two <strong>0-indexed</strong> strings <code>source</code> and <code>target</code>, both of length <code>n</code> and consisting of <strong>lowercase</strong> English letters. You are also given two <strong>0-indexed</strong> character arrays <code>original</code> and <code>changed</code>, and an integer array <code>cost</code>, where <code>cost[i]</code> represents the cost of changing the character <code>original[i]</code> to the character <code>changed[i]</code>.</p>

<p>You start with the string <code>source</code>. In one operation, you can pick a character <code>x</code> from the string and change it to the character <code>y</code> at a cost of <code>z</code> <strong>if</strong> there exists <strong>any</strong> index <code>j</code> such that <code>cost[j] == z</code>, <code>original[j] == x</code>, and <code>changed[j] == y</code>.</p>

<p>Return <em>the <strong>minimum</strong> cost to convert the string </em><code>source</code><em> to the string </em><code>target</code><em> using <strong>any</strong> number of operations. If it is impossible to convert</em> <code>source</code> <em>to</em> <code>target</code>, <em>return</em> <code>-1</code>.</p>

<p><strong>Note</strong> that there may exist indices <code>i</code>, <code>j</code> such that <code>original[j] == original[i]</code> and <code>changed[j] == changed[i]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> source = &quot;abcd&quot;, target = &quot;acbe&quot;, original = [&quot;a&quot;,&quot;b&quot;,&quot;c&quot;,&quot;c&quot;,&quot;e&quot;,&quot;d&quot;], changed = [&quot;b&quot;,&quot;c&quot;,&quot;b&quot;,&quot;e&quot;,&quot;b&quot;,&quot;e&quot;], cost = [2,5,5,1,2,20]
<strong>Output:</strong> 28
<strong>Explanation:</strong> To convert the string &quot;abcd&quot; to string &quot;acbe&quot;:
- Change value at index 1 from &#39;b&#39; to &#39;c&#39; at a cost of 5.
- Change value at index 2 from &#39;c&#39; to &#39;e&#39; at a cost of 1.
- Change value at index 2 from &#39;e&#39; to &#39;b&#39; at a cost of 2.
- Change value at index 3 from &#39;d&#39; to &#39;e&#39; at a cost of 20.
The total cost incurred is 5 + 1 + 2 + 20 = 28.
It can be shown that this is the minimum possible cost.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> source = &quot;aaaa&quot;, target = &quot;bbbb&quot;, original = [&quot;a&quot;,&quot;c&quot;], changed = [&quot;c&quot;,&quot;b&quot;], cost = [1,2]
<strong>Output:</strong> 12
<strong>Explanation:</strong> To change the character &#39;a&#39; to &#39;b&#39; change the character &#39;a&#39; to &#39;c&#39; at a cost of 1, followed by changing the character &#39;c&#39; to &#39;b&#39; at a cost of 2, for a total cost of 1 + 2 = 3. To change all occurrences of &#39;a&#39; to &#39;b&#39;, a total cost of 3 * 4 = 12 is incurred.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> source = &quot;abcd&quot;, target = &quot;abce&quot;, original = [&quot;a&quot;], changed = [&quot;e&quot;], cost = [10000]
<strong>Output:</strong> -1
<strong>Explanation:</strong> It is impossible to convert source to target because the value at index 3 cannot be changed from &#39;d&#39; to &#39;e&#39;.
</pre>

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

<ul>
	<li><code>1 &lt;= source.length == target.length &lt;= 10<sup>5</sup></code></li>
	<li><code>source</code>, <code>target</code> consist of lowercase English letters.</li>
	<li><code>1 &lt;= cost.length == original.length == changed.length &lt;= 2000</code></li>
	<li><code>original[i]</code>, <code>changed[i]</code> are lowercase English letters.</li>
	<li><code>1 &lt;= cost[i] &lt;= 10<sup>6</sup></code></li>
	<li><code>original[i] != changed[i]</code></li>
</ul>

ostream& operator<<(ostream& o, const pair<char, int>& neighbor) {
    o << "{" << neighbor.first << ", " << neighbor.second << "}";
    return o;
}

ostream& operator<<(ostream& o,
                    const vector<vector<pair<char, int>>>& adjList) {
    for (const auto& l : adjList) {
        for (const auto& neighbor : l) {
            o << neighbor << ", ";
        }
        o << endl;
    }
    return o;
};

class Solution {
    typedef vector<vector<pair<char, long>>> AdjList;
    void makeAdjList(const vector<char>& og, const vector<char>& changed,
                     const vector<int>& cost, AdjList& adj) {
        adj = AdjList(26, vector<pair<char, long>>());
        char letter = 'a', target = 'a';
        for (int i = 0; i < og.size(); i++) {
            letter = og[i];
            target = changed[i];
            adj[letter - 'a'].push_back(make_pair(target, cost[i]));
            ;
        }
    };

    typedef vector<vector<long>> Distance;
    void sssp(const AdjList& adj, Distance& dis) {
        // init Distance: i-th letter to i-th letter
        dis = Distance(26, vector<long>(26, -1));

        // pair: estimated cost, neighbor char
        priority_queue<pair<long, char>, vector<pair<long, char>>,
                       greater<pair<long, char>>>
            pQ;
        unordered_set<char> visited;
        long curr_cost = 0;
        for (int i = 0; i <= 'z' - 'a'; i++) {
            // Find all shortest path for letter i-th
            // Init. priority queue
            pQ = {};
            visited = {};
            pQ.emplace(0, i + 'a');
            curr_cost = 0;

            // While pQ is not empty, explore with Dijkstra
            while (!pQ.empty()) {
                auto [cost, neighbor] = pQ.top();
                pQ.pop();
                // verify if vertex is explored (part of the found shortest path
                // already)
                if (visited.contains(neighbor)) {
                    continue;
                }
                visited.insert(neighbor);

                // Record shortest path to this neighbor
                curr_cost = cost;
                dis[i][neighbor - 'a'] = curr_cost;
                //cout << (char)(i + 'a') << " to " << neighbor << ": " << curr_cost << endl;

                // Explore neighbor's adjacents
                for (const auto& [next, extra_cost] : adj[neighbor - 'a']) {
                    if (extra_cost + curr_cost < dis[i][next - 'a'] ||
                        dis[i][next - 'a'] == -1) {
                        // Shorter path found
                        pQ.emplace(extra_cost + curr_cost, next);
                    }
                }
            }
        }
    };

public:
    long long minimumCost(string source, string target, vector<char>& original,
                          vector<char>& changed, vector<int>& cost) {
        AdjList adjList;
        makeAdjList(original, changed, cost, adjList);
        Distance distance;
        sssp(adjList, distance);

        long total_cost = 0, subcost;
        for (int i = 0; i < source.size(); i++) {
            subcost = distance[source[i] - 'a'][target[i] - 'a'];
            //cout << "from " << source[i] << " to " <<target[i] << ": " << subcost << endl;;
            if (subcost == -1) {
                return -1;
            }
            total_cost += subcost;
        }

        return total_cost;
    }
};
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