<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>

class Solution {
private:
    void dji(vector<vector<pair<int, int>>> &edges, vector<vector<int>> &dist, int node) {
        vector<bool> visited(26, false);
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> que;
        que.push({0, node});

        while(!que.empty()) {
            pair<int, int> cur_node = que.top();
            que.pop();

            if (visited[cur_node.second]) continue;

            // cout << "node: " << cur_node.second << "\n";
            for (auto i : edges[cur_node.second]) {
                pair<int, int> new_node = {cur_node.first + i.first, i.second};
                que.push(new_node);
                // cout << "pushing: {" << new_node.second << "," << new_node.first << "} \n";
            }

            visited[cur_node.second] = true;
            dist[node][cur_node.second] = cur_node.first;
        }

        cout << endl;
    }
public:
    long long minimumCost(string source, string target, vector<char>& original, vector<char>& changed, vector<int>& cost) {
        int n = 26;
        vector<vector<pair<int, int>>> edges(n, vector<pair<int, int>>(0));
        vector<vector<int>> dist(n, vector<int>(n, -1));
        long long ans = 0;
        for (int i = 0; i < original.size(); i++) {
            int from = original[i] - 'a';
            int to = changed[i] - 'a';
            edges[from].push_back({cost[i], to});
        }

        for (int i = 0; i<n; i++) {
            dji(edges, dist, i);
        }

        // cout << "edges:\n";
        // for (int i = 0; i < 26; i++) {
        //     for (auto j : edges[i]){
        //         cout << j.second << " ";
        //     }
        //     cout << endl;
        // }

        // cout << "dist: \n";
        // int x = 10;
        // for (int i = 0; i<x; i++) {
        //     for (int j = 0; j<x; j++) {
        //         cout << dist[i][j] << " ";
        //     }
        //     cout << endl;
        // }

        for (int i = 0; i<source.size(); i++) {
            if (source[i] == target[i]) {
                continue;
            }

            int from = source[i] - 'a';
            int to = target[i] - 'a';

            if (dist[from][to] == -1) {
                return -1;
            }

            ans += dist[from][to];
        }

        return ans;
    }
};
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