<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>

vector<pair<int,double>> ad_list[55];
int vis[55];
double product;
int f;
void dfs(int u,int target,double cur)
{
    cout << u << " " << target << " " << cur << endl;
    if(u==target)  
    { 
        f = 1;
        product = cur;
        return;
    }
    vis[u] = 1;
    for(auto &i:ad_list[u])
    {
        if(!vis[i.first])
            dfs(i.first,target,cur*i.second);
    }
}
class Solution {
public:
    vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {
        for(int i = 0;i < 51;i++)
            ad_list[i].clear();
        map<string,int>mp;
        int x = 1;
        for(int i = 0;i < equations.size();i++)
        {
            string u = equations[i][0];
            string v = equations[i][1];
            int uu,vv;
            if(mp.count(u))
                uu = mp[u];
            else
                uu = x++,mp[u]=uu;
            if(mp.count(v))
                vv = mp[v];
            else
                vv = x++,mp[v]=vv;
            ad_list[uu].push_back({vv,values[i]});
            ad_list[vv].push_back({uu,1/values[i]});
        }
        vector<double>ans;
        for(int i = 0;i < queries.size();i++)
        {
            memset(vis,0,sizeof vis);
            string a = queries[i][0];
            string b = queries[i][1];
            int u = mp[a];
            int v = mp[b];
            // cout << a << " " << b << endl;
            if(u==0 or v == 0)
                ans.push_back(-1.00000);
            else
            {
                f = 0;
                dfs(u,v,1);
                if(f)
                    ans.push_back(product);
                else    
                    ans.push_back(-1);
            }
        }
        // dbg(ans);
        return ans;
    }
};
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