<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
{
    class FloydWarshall
    {
    private:
        vector<vector<int>> distance;
        vector<vector<int>> next;

        int nrVertex;

    public:
        FloydWarshall(int n)
        {
            nrVertex = n;

            distance.resize(n);
            next.resize(n);

            for (int i = 0; i < n; ++i)
            {
                distance[i].resize(n, 999999999);
                next[i].resize(n, -1);
            }

            for (int i = 0; i < n; i++)
            {
                distance[i][i] = 0;
                next[i][i] = i;
            }
        }

        void AddVertex(int origin, int destiny, int cost, bool bidirectional)
        {
            if (cost < distance[origin][destiny])
            {
                distance[origin][destiny] = cost;
                next    [origin][destiny] = destiny;
            }

            if (bidirectional)
            {
                if (cost < distance[destiny][origin])
                {
                    distance[destiny][origin] = cost;
                    next    [destiny][origin] = origin;
                }
            }
        }

        void Execute()
        {
            for (int k = 0; k < nrVertex; k++)
            {
                for (int i = 0; i < nrVertex; i++)
                {
                    for (int j = 0; j < nrVertex; j++)
                    {
                        if (distance[i][j] > distance[i][k] + distance[k][j])
                        {
                            distance[i][j] = distance[i][k] + distance[k][j];
                            next[i][j] = next[i][k];
                        }
                    }
                }
            }
        }

        int GetCost(int origin, int destiny)
        {
            return distance[origin][destiny];
        }

        bool HasPath(int origin, int destiny)
        {
            return next[origin][destiny] != -1;
        }
    };


public:
    long long minimumCost(string source, string target, vector<char>& original, vector<char>& changed, vector<int>& cost)
    {
        FloydWarshall fw(26);
        for (int i = 0; i < original.size(); ++i)
            fw.AddVertex(original[i] - 97, changed[i] - 97, cost[i], false);

        fw.Execute();

        long long sum = 0;
        for (int i = 0; i < source.size(); ++i)
        {
            if (!fw.HasPath(source[i] - 97, target[i] - 97))
            {
                sum = -1;
                break;
            }

            sum += fw.GetCost(source[i] - 97, target[i] - 97);
        }

        return sum;
    }
};

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