Algorithms（三）Evaluate Division

題目：

Evaluate Division

Equations are given in the format A / B = k, where A andB are variables represented as strings, andk 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, whereequations.size() == values.size(), and the values are positive. This represents the equations. Returnvector<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.

---

代碼：

class Solution {
public:
    vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {
        // 註釋1
        unordered_map<string, Node*> map; // 註釋2
        vector<double> res;
        for (int i = 0; i < equations.size(); i++) {
            string s1 = equations[i].first, s2 = equations[i].second;
            if (map.count(s1) == 0 && map.count(s2) == 0) {
                map[s1] = new Node();
                map[s2] = new Node();
                map[s1] -> value = values[i];
                map[s2] -> value = 1;
                map[s1] -> parent = map[s2];
            } else if (map.count(s1) == 0) {
                map[s1] = new Node();
                map[s1] -> value = map[s2] -> value * values[i];
                map[s1] -> parent = map[s2];
            } else if (map.count(s2) == 0) {
                map[s2] = new Node();
                map[s2] -> value = map[s1] -> value / values[i];
                map[s2] -> parent = map[s1];
            } else {
                unionNodes(map[s1], map[s2], values[i], map);
            }
        }

        for (auto query : queries) {
            if (map.count(query.first) == 0 || map.count(query.second) == 0 || findParent(map[query.first]) !=
                findParent(map[query.second]))
                res.push_back(-1);
            else
                res.push_back(map[query.first] -> value / map[query.second] -> value);
        }
        return res;
    }
    
private:
    struct Node {
        Node* parent;
        double value = 0.0;
        Node()  {parent = this;}
    };
    
    void unionNodes(Node* node1, Node* node2, double num, unordered_map<string, Node*>& map) {
        Node* parent1 = findParent(node1), *parent2 = findParent(node2);
        double ratio = node2 -> value * num / node1 -> value;
        for (auto it = map.begin(); it != map.end(); it++) { // 註釋3
            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;
    }
};

算法：

1. 遍歷equations的每一項，第i個等式滿足 s1/s2 = values[i]，分四種情況討論：

· 若圖中沒有s1，s2，則建立兩個新節點，令s2 = 1，s1 = values[i]。因爲s1是以s2的值爲基準取值的，所以令s1的父節點爲s2。

· 若圖中沒有s1，有s2，則建立新節點s1，令s1 = s2 * values[i]。因爲s1是以s2的值爲基準取值的，所以令s1的父節點爲s2。

· 若圖中有s1，沒有s2，則建立新節點s2，令s2 = s1 / values[i]。因爲此時，s2是以s1的值爲基準進行取值的，所以令s2的父節點爲s1。

· 若圖中已經有了s1，s2，則從s1的父節點不斷遞歸，找到s1的根節點parent1，同理，找到s2的根節點。然後將圖中所有根節點爲parent1的點（包括s1的根節點本身），都乘以參數ratio，該參數滿足s1 * ratio / s2 = values[i]。因爲此時，parent1是以parent2爲基準取值的，所以令parent1的父節點爲parent2。

2. 遍歷queries的每一項，分兩種情況處理：

· 若queries中的兩個量s1、s2沒有同時存在與圖中，或者這兩個量的根節點不同，無法比較，則返回-1.

· 若s1/s2的值存在，則返回該值。

註釋：

註釋1：c++ 中 pair 的 使用方法

註釋2：C++ map的基本操作和使用 ； Map集合的四種遍歷方式            

註釋3：C++11特性：auto關鍵字；C++ iterator->second意思