Algorithms（三）Evaluate Division

题目：

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.

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代码：

class Solution {public:    vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {        // 注释1        unordered_map<string, Node*> map; // 注释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++) { // 注释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. 遍历equations的每一项，第i个等式满足 s1/s2 = values[i]，分四种情况讨论：

· 若图中没有s1，s2，则建立两个新节点，令s2 = 1，s1 = values[i]。因为s1是以s2的值为基准取值的，所以令s1的父节点为s2。

· 若图中没有s1，有s2，则建立新节点s1，令s1 = s2 * values[i]。因为s1是以s2的值为基准取值的，所以令s1的父节点为s2。

· 若图中有s1，没有s2，则建立新节点s2，令s2 = s1 / values[i]。因为此时，s2是以s1的值为基准进行取值的，所以令s2的父节点为s1。

· 若图中已经有了s1，s2，则从s1的父节点不断递归，找到s1的根节点parent1，同理，找到s2的根节点。然后将图中所有根节点为parent1的点（包括s1的根节点本身），都乘以参数ratio，该参数满足s1 * ratio / s2 = values[i]。因为此时，parent1是以parent2为基准取值的，所以令parent1的父节点为parent2。

2. 遍历queries的每一项，分两种情况处理：

· 若queries中的两个量s1、s2没有同时存在与图中，或者这两个量的根节点不同，无法比较，则返回-1.

· 若s1/s2的值存在，则返回该值。

注释：

注释1：c++ 中 pair 的 使用方法

注释2：C++ map的基本操作和使用 ； Map集合的四种遍历方式            

注释3：C++11特性：auto关键字；C++ iterator->second意思