Minimum Cost to Convert String II
<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 characters. You are also given two <strong>0-indexed</strong> string arrays <code>original</code> and <code>changed</code>, and an integer array <code>cost</code>, where <code>cost[i]</code> represents the cost of converting the string <code>original[i]</code> to the string <code>changed[i]</code>.</p>

<p>You start with the string <code>source</code>. In one operation, you can pick a <strong>substring</strong> <code>x</code> from the string, and change it to <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>. You are allowed to do <strong>any</strong> number of operations, but any pair of operations must satisfy <strong>either</strong> of these two conditions:</p>

<ul>
	<li>The substrings picked in the operations are <code>source[a..b]</code> and <code>source[c..d]</code> with either <code>b &lt; c</code> <strong>or</strong> <code>d &lt; a</code>. In other words, the indices picked in both operations are <strong>disjoint</strong>.</li>
	<li>The substrings picked in the operations are <code>source[a..b]</code> and <code>source[c..d]</code> with <code>a == c</code> <strong>and</strong> <code>b == d</code>. In other words, the indices picked in both operations are <strong>identical</strong>.</li>
</ul>

<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</em>. <em>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 &quot;abcd&quot; to &quot;acbe&quot;, do the following operations:
- Change substring source[1..1] from &quot;b&quot; to &quot;c&quot; at a cost of 5.
- Change substring source[2..2] from &quot;c&quot; to &quot;e&quot; at a cost of 1.
- Change substring source[2..2] from &quot;e&quot; to &quot;b&quot; at a cost of 2.
- Change substring source[3..3] from &quot;d&quot; to &quot;e&quot; 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;abcdefgh&quot;, target = &quot;acdeeghh&quot;, original = [&quot;bcd&quot;,&quot;fgh&quot;,&quot;thh&quot;], changed = [&quot;cde&quot;,&quot;thh&quot;,&quot;ghh&quot;], cost = [1,3,5]
<strong>Output:</strong> 9
<strong>Explanation:</strong> To convert &quot;abcdefgh&quot; to &quot;acdeeghh&quot;, do the following operations:
- Change substring source[1..3] from &quot;bcd&quot; to &quot;cde&quot; at a cost of 1.
- Change substring source[5..7] from &quot;fgh&quot; to &quot;thh&quot; at a cost of 3. We can do this operation because indices [5,7] are disjoint with indices picked in the first operation.
- Change substring source[5..7] from &quot;thh&quot; to &quot;ghh&quot; at a cost of 5. We can do this operation because indices [5,7] are disjoint with indices picked in the first operation, and identical with indices picked in the second operation.
The total cost incurred is 1 + 3 + 5 = 9.
It can be shown that this is the minimum possible cost.
</pre>

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

<pre>
<strong>Input:</strong> source = &quot;abcdefgh&quot;, target = &quot;addddddd&quot;, original = [&quot;bcd&quot;,&quot;defgh&quot;], changed = [&quot;ddd&quot;,&quot;ddddd&quot;], cost = [100,1578]
<strong>Output:</strong> -1
<strong>Explanation:</strong> It is impossible to convert &quot;abcdefgh&quot; to &quot;addddddd&quot;.
If you select substring source[1..3] as the first operation to change &quot;abcdefgh&quot; to &quot;adddefgh&quot;, you cannot select substring source[3..7] as the second operation because it has a common index, 3, with the first operation.
If you select substring source[3..7] as the first operation to change &quot;abcdefgh&quot; to &quot;abcddddd&quot;, you cannot select substring source[1..3] as the second operation because it has a common index, 3, with the first operation.
</pre>

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

<ul>
	<li><code>1 &lt;= source.length == target.length &lt;= 1000</code></li>
	<li><code>source</code>, <code>target</code> consist only of lowercase English characters.</li>
	<li><code>1 &lt;= cost.length == original.length == changed.length &lt;= 100</code></li>
	<li><code>1 &lt;= original[i].length == changed[i].length &lt;= source.length</code></li>
	<li><code>original[i]</code>, <code>changed[i]</code> consist only of lowercase English characters.</li>
	<li><code>original[i] != changed[i]</code></li>
	<li><code>1 &lt;= cost[i] &lt;= 10<sup>6</sup></code></li>
</ul>
Graph; Trie; Array; String; Dynamic Programming; Shortest Path
class Node {
public:
    Node* children[26];
    int v = -1;
    Node() {
        fill(children, children + 26, nullptr);
    }
};

class Solution {
private:
    const long long inf = 1LL << 60;
    Node* root = new Node();
    int idx;

    vector<vector<long long>> g;
    string s;
    string t;
    vector<long long> f;

public:
    long long minimumCost(string source, string target, vector<string>& original, vector<string>& changed, vector<int>& cost) {
        int m = cost.size();
        g = vector<vector<long long>>(m << 1, vector<long long>(m << 1, inf));
        s = source;
        t = target;

        for (int i = 0; i < g.size(); ++i) {
            g[i][i] = 0;
        }

        for (int i = 0; i < m; ++i) {
            int x = insert(original[i]);
            int y = insert(changed[i]);
            g[x][y] = min(g[x][y], static_cast<long long>(cost[i]));
        }

        for (int k = 0; k < idx; ++k) {
            for (int i = 0; i < idx; ++i) {
                if (g[i][k] >= inf) {
                    continue;
                }
                for (int j = 0; j < idx; ++j) {
                    g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
                }
            }
        }

        f = vector<long long>(s.length(), -1);
        long long ans = dfs(0);
        return ans >= inf ? -1 : ans;
    }

private:
    int insert(const string& w) {
        Node* node = root;
        for (char c : w) {
            int i = c - 'a';
            if (node->children[i] == nullptr) {
                node->children[i] = new Node();
            }
            node = node->children[i];
        }
        if (node->v < 0) {
            node->v = idx++;
        }
        return node->v;
    }

    long long dfs(int i) {
        if (i >= s.length()) {
            return 0;
        }
        if (f[i] != -1) {
            return f[i];
        }
        long long res = (s[i] == t[i]) ? dfs(i + 1) : inf;
        Node* p = root;
        Node* q = root;
        for (int j = i; j < s.length(); ++j) {
            p = p->children[s[j] - 'a'];
            q = q->children[t[j] - 'a'];
            if (p == nullptr || q == nullptr) {
                break;
            }
            if (p->v < 0 || q->v < 0) {
                continue;
            }
            long long temp = g[p->v][q->v];
            if (temp < inf) {
                res = min(res, temp + dfs(j + 1));
            }
        }
        return f[i] = res;
    }
};