中山大学算法课程题目详解（第五周）

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.

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.

问题解决思路：

首先，主要到每个除法表达式都有个值，而且是双向的，可以用图的方法来解决，这里要特别注意图是无向图，例如等式a / b = 2.0，可以转化为两条边：<a, b>，<b, a>，其长度分别为2.0，0.5。先判断2个点是否都出现，然后判断是不是自己到自己，最后用dfs进行搜索，然后用递归进行每条边权重的相乘，如果最后找不到路径，则返回-1.0；这里采用嵌套map作为存储对应边的两个点以及权值，采用set存储在dfs中以及被访问的点。

具体实现代码：

class Solution {public:    double dfs(string from, string to, map<string, map<string, double>> &mp, set<string> & visited, double val){if (mp[from].count(to)) {return mp[from][to] * val;}for (pair<string, double>p : mp[from]) {string str = p.first;if (visited.count(str) == 0) {visited.insert(str);double cur = dfs(str, to, mp, visited, p.second * val);if (cur != -1.0) return cur;}}return -1.0;}vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {vector<double> result;map<string, map<string, double>> mp;int i = 0;for (pair<string, string> p : equations) {mp[p.first][p.second] = values[i];mp[p.second][p.first] = 1.0 / values[i];i++;}for (pair<string, string> p : queries) {if (mp.count(p.first) && mp.count(p.second)) {if (p.first == p.second) {result.push_back(1.0);continue;}else {set<string> visited;result.push_back(dfs(p.first, p.second, mp, visited, 1.0));}}else {result.push_back(-1.0);}}return result;}};