Evaluate Division问题及解法

问题描述：

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.

示例：
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.

问题分析：

此题是图遍历问题，对此我们先构建邻接表，然后利用dfs遍历图结构即可得到答案。

过程详见代码：

class Solution {public:    vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {unordered_map<string, vector<pair<string, double>>> map;vector<double> res;unordered_map<string, int> used;for (int i = 0; i < equations.size(); i++) // 构造邻接表{map[equations[i].first].emplace_back(pair<string, double>(equations[i].second, values[i]));map[equations[i].second].emplace_back(pair<string, double>(equations[i].first, 1 / values[i]));used[equations[i].first] = 0;used[equations[i].second] = 0;}for (int i = 0; i < queries.size(); i++){double r = -1.0;if (map.count(queries[i].first) && map.count(queries[i].second)){used[queries[i].first] = 1;dfs(queries[i].first, queries[i].second, map, r, 1, used);used[queries[i].first] = 0;}res.emplace_back(r);}return res;}void dfs(string first, string second, unordered_map<string, vector<pair<string, double>>> map,double& r,double val,unordered_map<string,int>& used){if (first == second){r = val;return;}vector<pair<string, double>> vsd = map[first];for (int i = 0; i < vsd.size(); i++){if (r == -1.0){if (!used[vsd[i].first]){ used[vsd[i].first] = 1;dfs(vsd[i].first, second, map, r, val * vsd[i].second, used);used[vsd[i].first] = 0;}}else return;}}};