<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
{
    int n, nb;
    string source, target;
    map<char, int> ids;
    vector<map<int, long long>> graph;
    vector<long long> cache;
    long long solve(int pos)
    {
        if (pos == n)
            return 0;
        if (cache[pos] != -2)
            return cache[pos];
        long long res = -1;
        if (source[pos] == target[pos])
            res = solve(pos + 1);
        for (const auto& [s, id] : ids)
        {
            if (source[pos] != s)
                continue;
            auto it = ids.find(target[pos]);
            if (it == ids.end())
                continue;
            auto it2 = graph[id].find(it->second);
            if (it2 == graph[id].end())
                continue;
            long long remain = solve(pos + 1);
            if (remain != -1 && (res == -1 || remain + it2->second < res))
                res = remain + it2->second;
        }
        cache[pos] = res;
        return res;
    }
  public:
    long long minimumCost(string source, string target, vector<char>& original, vector<char>& changed, vector<int>& cost)
    {
        this->source = source;
        this->target = target;
        n = source.size();
        for (int i = 0; i < original.size(); ++i)
        {
            auto ito = ids.find(original[i]);
            int ido = -1;
            if (ito == ids.end())
            {
                ido = graph.size();
                ids[original[i]] = ido;
                graph.push_back({});
            }
            else
                ido = ito->second;
            auto itc = ids.find(changed[i]);
            int idc = -1;
            if (itc == ids.end())
            {
                idc = graph.size();
                ids[changed[i]] = idc;
                graph.push_back({});
            }
            else
                idc = itc->second;
            if (graph[ido][idc] == 0)
                graph[ido][idc] = cost[i];
            else
                graph[ido][idc] = min(graph[ido][idc], (long long)cost[i]);
        }
        nb = graph.size();
        for (int i = 0; i < nb; ++i)
        {
            vector<pair<long long, int>> heap;
            for (auto [j, d] : graph[i])
            {
                heap.push_back({-d, j});
                push_heap(heap.begin(), heap.end());
            }
            while (!heap.empty())
            {
                pop_heap(heap.begin(), heap.end());
                auto [d1, j] = heap.back();
                d1 = -d1;
                heap.pop_back();
                if (auto it = graph[i].find(j); it != graph[i].end() && it->second < d1)
                    continue;
                for (auto [k, d2] : graph[j])
                {
                    if (auto it = graph[i].find(k); it == graph[i].end() || it->second > d1 + d2)
                    {
                        graph[i][k] = d1 + d2;
                        heap.push_back({-d1 - d2, k});
                        push_heap(heap.begin(), heap.end());
                    }
                }
            }
        }
        cache = vector<long long>(n, -2);
        return solve(0);
    }
};

139958