题目来源
给一堆等式，求另一堆等式的值。
用并查集来做，然后我不熟悉，边看讨论区边写的…写完觉得实际上也不难，就是看着有点长而已。
代码如下：

class Solution {
public:
    vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {
        unordered_map<string, Node*> maps;
        vector<double> res;
        for (int i=0; i<equations.size(); i++) {
            string s1 = equations[i].first, s2 = equations[i].second;
            if (maps.count(s1) == 0 && maps.count(s2) == 0) {
                maps[s1] = new Node();
                maps[s2] = new Node();
                maps[s2]->value = 1;
                maps[s1]->value = values[i];
                maps[s2]->parent = maps[s1];
            }
            else if (maps.count(s1) == 0) {
                maps[s1] = new Node();
                maps[s1]->value = maps[s2]->value * values[i];
                maps[s1]->parent = maps[s2];
            }
            else if (maps.count(s2) == 0) {
                maps[s2] = new Node();
                maps[s2]->value = maps[s1]->value / values[i];
                maps[s2]->parent = maps[s1];
            }
            else {
                unionNode(maps[s1], maps[s2], values[i], maps);
            }
        }
        
        for (auto query : queries) {
            if (maps.count(query.first) == 0 || maps.count(query.second) == 0 || 
            findParent(maps[query.first]) != findParent(maps[query.second]))
                res.push_back(-1);
            else
                res.push_back(maps[query.first]->value / maps[query.second]->value);
        }
        return res;
    }
    
private:
    struct Node {
        Node* parent;
        double value = 0;
        Node() {parent = this;}
    };
    
    void unionNode(Node* node1, Node* node2, double num, unordered_map<string, Node*>& maps)
    {
        Node* parent1 = findParent(node1), *parent2 = findParent(node2);
        double ratio = node2->value * num / node1->value;
        for (auto it=maps.begin(); it!=maps.end(); it++)
            if (findParent(it->second) == parent1)
                it->second->value *= ratio;
        parent1->parent = parent2;
    }
    
    Node* findParent(Node* node)
    {
        if (node->parent == node)
            return node;
        node->parent = findParent(node->parent);
        return node->parent;
    }
};