leetcode 399. Evaluate Division

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本文介绍了一个基于图的搜索算法来解决EvaluateDivision问题的方法。通过构建一个图，利用BFS算法来寻找两个变量之间的关系路径，并计算其乘积值。若路径不存在，则返回-1.0。适用于给定方程组求解未知比值问题。

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<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.

思路就是 a/c = a/b * b/c 所以其实是一个搜索路径的问题。

a是start，c是end。看能不能从start搜到end。

所以用标准BFS来做。

class Solution {
public:
    vector<double> calcEquation(vector<pair<string, string>> equ, vector<double>& values, vector<pair<string, string>> que) 
    {
        vector<double> ret;  //返回的结果
        map<string, set<pair<string, double>>> mp; //BFS使用
        
        for (int i = 0; i < equ.size(); i++)    //正反 和 本身 都存进去
        {
            mp[equ[i].first].insert(make_pair(equ[i].first, 1.0));
            mp[equ[i].second].insert(make_pair(equ[i].second, 1.0));
            mp[equ[i].first].insert(make_pair(equ[i].second, values[i])); 
            mp[equ[i].second].insert(make_pair(equ[i].first, 1 / values[i])); 
        }
        
        for (int i = 0; i < que.size(); i++)    
        {
            string start = que[i].first;
            string end = que[i].second;
            
            //排除 [x , x] 等从未出现过的情况
            if (mp.find(start) == mp.end() || mp.find(end) == mp.end())
            {
                ret.push_back(-1.0);
                continue;
            }
            
            int flag = 0;//用来判断是否有答案，没有的话是 -1
            
            queue<pair<string, double>> route;  //BFS
            set<string> hash;  //BFS搜索过的存下来 防止再次搜索
            hash.insert(start); 
            route.push(make_pair(start, 1.0));
            while (!route.empty())
            {
                string now_str = route.front().first;
                double now_v = route.front().second;
                route.pop();
                
                //结束条件
                if (now_str == end)
                {
                    ret.push_back(now_v);
                    flag = 1;
                    break;
                }

                for (auto it : mp[now_str]) //BFS
                {
                    if (hash.find(it.first) == hash.end() )  //本身就不再进入 BFS 搜索队列了
                    {
                        route.push(make_pair(it.first, it.second * now_v));
                        hash.insert(it.first);
                    }
                } 
            }
            if (flag == 0)
                ret.push_back(-1.0);
        }
        return ret;    
    }
};