Evaluate Division -- LeetCode 399

[Question]

Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0.

Example:
Given a / b = 2.0, b / c = 3.0.
queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? .
return [6.0, 0.5, -1.0, 1.0, -1.0 ].

The input is: vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries , where equations.size() == values.size(), and the values are positive. This represents the equations. Return vector<double>.

According to the example above:

equations = [ ["a", "b"], ["b", "c"] ],
values = [2.0, 3.0],
queries = [ ["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"] ]. 

The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction.

[Analysis]
Consider each equation as an edge in a directed graph, each string is a vertex, then the problem becomes to find a path for each pair of strings in the queries array.

Inspired by Floyd-Warshall algorithm, using a 2-D matrix to represent vertex A to vertex path (if exists), A[i][j] = A[i][k]*A[k][j], for k=0,...|v|-1.

[Solution]
class Solution {
public:
    vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {
        set<string> nodes;
        unordered_map<string, int> inv;
     
        for (auto& e:equations) {
            nodes.insert(e.first);
            nodes.insert(e.second);
        }
        int i=0;
        for (auto it= nodes.begin(); it!=nodes.end(); it++, i++)
            inv[*it] = i;
     
        vector<vector<double>> equ(nodes.size(), vector<double>(nodes.size(),-1.0));
        for (int i=0; i< nodes.size(); i++)
            equ[i][i]= 1.0;
         
        for (int i=0; i< equations.size(); i++) {
            int x = inv[equations[i].first];
            int y = inv[equations[i].second];
            equ[x][y] = values[i];
            equ[y][x] = 1.0 / values[i];
        }
     
        for (int k=0; k<nodes.size(); k++) {
            for (int i=0; i<nodes.size(); i++) {
                for (int j=i+1; j<nodes.size(); j++) {
                    if (equ[i][k]!=-1.0 && equ[k][j]!=-1.0) {
                        equ[i][j] = equ[i][k] * equ[k][j];
                        equ[j][i] = 1.0/ equ[i][j];
                    }
                }
            }
        }
     
        vector<double> res;
        for (auto& q: queries) {
            if (nodes.count(q.first) && nodes.count(q.second)) {
                int x= inv[q.first], y= inv[q.second];
                res.push_back( equ[x][y] );
            }
            else res.push_back(-1.0);
        }
        return res;
    }

};