[leetcode] 399. Evaluate Division 解題報告

題目鏈接：https://leetcode.com/problems/evaluate-division/

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 , whereequations.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.

思路：形如給定a/b和b/c，讓求a/c，可以抽象成圖論求各結點之間的距離．因爲是多點到多點的方式，所以可以用Floyd 算法來算．這個算法應該是圖論中最簡單粗暴的算法了，相比prime和dijkstra算法不僅容易理解，還很容易實現．只要三重循環即可，其中最外層是中間結點．願一切都如此算法一樣簡單．還有要注意的是如果queries的是兩個相同數如a/a這種自己設置一下即可．

代碼如下：

class Solution {
public:
    vector<double> calcEquation(vector<pair<string, string>> equations, 
        vector<double>& values, vector<pair<string, string>> queries) {
        unordered_map<string, unordered_map<string, double>> hash;
        for(int i = 0; i < equations.size(); i++) 
        {
            hash[equations[i].first][equations[i].second] = values[i];
            hash[equations[i].second][equations[i].first] = 1/values[i];
        }
        for(auto val: hash) hash[val.first][val.first] = 1;
        for(auto val1: hash)
        {
            for(auto val2: hash)
                for(auto val3: hash)
                    if(hash[val1.first].count(val3.first) 
                            && hash[val2.first].count(val1.first))
                        hash[val2.first][val3.first] = 
                            hash[val2.first][val1.first]*hash[val1.first][val3.first];
        }
        vector<double> ans;
        for(auto val: queries)
            ans.push_back(hash[val.first].count(val.second)?hash[val.first][val.second]:-1);
        return ans;
    }
};