<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:
    unordered_map<string, unordered_map<string, double>> graph;
    unordered_map<string, unordered_map<string, double>> memo; // 缓存中间结果
public:
    vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {
        // 想像成grap
        // a / b = a -> b
        // a -> c = a -> b -> c =  (a / b) * (b / c)
        // a / b 倒過來可以得到 b -> a

        // create graph
        for (int i = 0; i < equations.size(); i++) {
            graph[equations[i][0]][equations[i][1]] = values[i];
            graph[equations[i][1]][equations[i][0]] = 1.0 / values[i];
        }

        // DFS traversal
        vector<double> ret(queries.size());
        for (int i = 0; i < queries.size(); i++) {
            string u = queries[i][0];
            string v = queries[i][1];
            // u跟v判斷一次就可以，所以不需要寫在dfs中
            if (graph.find(u) == graph.end() || graph.find(v) == graph.end()) {
                ret[i] = -1.0;
            } else if (u == v) {
                ret[i] = 1.0;
            } else {
                unordered_set<string> visited;
                ret[i] = DFS(u, v, visited);
            }
        }

        return ret;
    }
    double DFS(string u, string v, unordered_set<string>& visited) {

        // 如果缓存中已有结果，直接返回
        if (memo.find(u) != memo.end() && memo[u].find(v) != memo[u].end()) {
            return memo[u][v];
        }
        if (graph[u].find(v) != graph[u].end()) {
            memo[u][v] = graph[u][v];
            return graph[u][v];
        }

        visited.insert(u);

        for (auto neighbor: graph[u]) {
            // 確保還沒visited過
            if (visited.find(neighbor.first) == visited.end()) {
                double weight = DFS(neighbor.first, v, visited);
                if (weight != -1) {
                    memo[u][v] = weight * neighbor.second;  // 缓存结果
                    return memo[u][v];
                }
            }
        }

        // 未找到结果，缓存结果并返回 -1
        //memo[u][v] = -1.0;

        // not found
        return -1.0;
    }
};
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