<p>You are given an array of variable pairs <code>equations</code> and an array of real numbers <code>values</code>, where <code>equations[i] = [A<sub>i</sub>, B<sub>i</sub>]</code> and <code>values[i]</code> represent the equation <code>A<sub>i</sub> / B<sub>i</sub> = values[i]</code>. Each <code>A<sub>i</sub></code> or <code>B<sub>i</sub></code> is a string that represents a single variable.</p>

<p>You are also given some <code>queries</code>, where <code>queries[j] = [C<sub>j</sub>, D<sub>j</sub>]</code> represents the <code>j<sup>th</sup></code> query where you must find the answer for <code>C<sub>j</sub> / D<sub>j</sub> = ?</code>.</p>

<p>Return <em>the answers to all queries</em>. If a single answer cannot be determined, return <code>-1.0</code>.</p>

<p><strong>Note:</strong> The input is always valid. You may assume that evaluating the queries will not result in division by zero and that there is no contradiction.</p>

<p><strong>Note:&nbsp;</strong>The variables that do not occur in the list of equations are undefined, so the answer cannot be determined for them.</p>

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

<pre>
<strong>Input:</strong> equations = [[&quot;a&quot;,&quot;b&quot;],[&quot;b&quot;,&quot;c&quot;]], values = [2.0,3.0], queries = [[&quot;a&quot;,&quot;c&quot;],[&quot;b&quot;,&quot;a&quot;],[&quot;a&quot;,&quot;e&quot;],[&quot;a&quot;,&quot;a&quot;],[&quot;x&quot;,&quot;x&quot;]]
<strong>Output:</strong> [6.00000,0.50000,-1.00000,1.00000,-1.00000]
<strong>Explanation:</strong> 
Given: <em>a / b = 2.0</em>, <em>b / c = 3.0</em>
queries are: <em>a / c = ?</em>, <em>b / a = ?</em>, <em>a / e = ?</em>, <em>a / a = ?</em>, <em>x / x = ? </em>
return: [6.0, 0.5, -1.0, 1.0, -1.0 ]
note: x is undefined =&gt; -1.0</pre>

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

<pre>
<strong>Input:</strong> equations = [[&quot;a&quot;,&quot;b&quot;],[&quot;b&quot;,&quot;c&quot;],[&quot;bc&quot;,&quot;cd&quot;]], values = [1.5,2.5,5.0], queries = [[&quot;a&quot;,&quot;c&quot;],[&quot;c&quot;,&quot;b&quot;],[&quot;bc&quot;,&quot;cd&quot;],[&quot;cd&quot;,&quot;bc&quot;]]
<strong>Output:</strong> [3.75000,0.40000,5.00000,0.20000]
</pre>

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

<pre>
<strong>Input:</strong> equations = [[&quot;a&quot;,&quot;b&quot;]], values = [0.5], queries = [[&quot;a&quot;,&quot;b&quot;],[&quot;b&quot;,&quot;a&quot;],[&quot;a&quot;,&quot;c&quot;],[&quot;x&quot;,&quot;y&quot;]]
<strong>Output:</strong> [0.50000,2.00000,-1.00000,-1.00000]
</pre>

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

<ul>
	<li><code>1 &lt;= equations.length &lt;= 20</code></li>
	<li><code>equations[i].length == 2</code></li>
	<li><code>1 &lt;= A<sub>i</sub>.length, B<sub>i</sub>.length &lt;= 5</code></li>
	<li><code>values.length == equations.length</code></li>
	<li><code>0.0 &lt; values[i] &lt;= 20.0</code></li>
	<li><code>1 &lt;= queries.length &lt;= 20</code></li>
	<li><code>queries[i].length == 2</code></li>
	<li><code>1 &lt;= C<sub>j</sub>.length, D<sub>j</sub>.length &lt;= 5</code></li>
	<li><code>A<sub>i</sub>, B<sub>i</sub>, C<sub>j</sub>, D<sub>j</sub></code> consist of lower case English letters and digits.</li>
</ul>

class Solution {
private:
    double dfs(unordered_map<string, unordered_map<string, double>>& map, string& source, string& dest, set<pair<string, string>>& visited) {
        if(source == dest) return 1.0;
        if(map.find(source) == map.end()) return -1.0;
        for(auto it: map[source]) {
            string next = it.first;
            if(visited.find({source, next}) != visited.end()) continue;
            visited.insert({source, next});
            double val = dfs(map, next, dest, visited);
            if(val != -1) return map[source][next] * val;
        }
        return -1.0;
    }

public:
    vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {
        int n = equations.size();
        unordered_map<string, unordered_map<string, double>> map;
        unordered_set<string> seen;
        for(int i=0; i<n; i++) {
            string a = equations[i][0], b = equations[i][1];
            seen.insert(a);
            seen.insert(b);
            double val = values[i];
            if(map.find(a) == map.end()) map[a] = unordered_map<string, double>();
            map[a][b] = val;
            if(val != 0.0) {
                if(map.find(b) == map.end()) map[b] = unordered_map<string, double>();
                map[b][a] = 1/val;
            }
        }
        int qsize = queries.size();
        vector<double> res(qsize);
        for(int i=0; i<qsize; i++) {
            string qa = queries[i][0], qb = queries[i][1];
            if(seen.find(qa) == seen.end() || seen.find(qb) == seen.end()) {
                res[i] = -1.0;
                continue;
            }
            if(qa == qb) {
                res[i] = 1.0;
                continue;
            }
            set<pair<string, string>> visited;
            res[i] = dfs(map, qa, qb, visited);
        }
        return res;
    }
};
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