399. Evaluate Division

399. 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

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.

Analyze:
从题目分析，本题可以采用图论的方法。首先看到除法规则就会想到除0的问题，因此在构造图的过程中，若结果为0的时候，路径是单向，否则路径是双向的。在构造好图以后，只需要对查询的字符串做深度优先搜索即可。

Code:

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<unordered_set>
#include<unordered_map>

using namespace std;

class Solution {
public:
    vector<double> calcEquation(vector<pair<string, string>> equations,
        vector<double>& values, vector<pair<string, string>> query)
    {
        unordered_map<string, unordered_map<string, double>> m;
        vector<double> res;
        // 创造有向图
        for (int i = 0; i < values.size(); ++i) {
            m[equations[i].first].insert(make_pair(equations[i].second, values[i]));
            if (values[i] != 0)
                m[equations[i].second].insert(make_pair(equations[i].first, 1 / values[i]));
        }

        for (auto i : query)
        {
            unordered_set<string> s;
            double tmp = check(i.first, i.second, m, s);
            if (tmp) res.push_back(tmp);
            else res.push_back(-1);
        }
        return res;
    }

    double check(string up, string down,
        unordered_map<string, unordered_map<string, double>> &m,
        unordered_set<string> &s)
    {
        if (m[up].find(down) != m[up].end()) return m[up][down];
        for (auto i : m[up])
        {
            if (s.find(i.first) == s.end())
            {
                s.insert(i.first);
                double tmp = check(i.first, down, m, s);
                if (tmp) return i.second*tmp;
            }
        }
        return 0;
    }
};