LN : leetcode 399 Evaluate Division

lc 399 Evaluate Division

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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.

DFS Accepted##

这个星期老师讲了图的分解以及DFS算法，所以我就在leetcode上找了一道相关的题目。问题的关键在于创建一个有向图，对于每一个字符串结点来说，应用map<string, map<string,double>> node这样的数据结构来存储每个有向边是很适应该问题的。根据equations和values中的值来创建有向边，正向为values[i]，反向为1/values[i]。

创建完有向图后，利用DFS对每个query从起点出发进行搜索，每次都乘以该条路径上的值，若两点间不能连通或点不在node范围之内，则返回0，最终输出-1.0。

class Solution {
public:
    vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {
        vector<double> ans;
        map<string, map<string,double>> node;
        for (int i = 0; i < values.size(); i++) {
            node[equations[i].first].insert(make_pair(equations[i].second, values[i]));
            node[equations[i].second].insert(make_pair(equations[i].first, 1/values[i]));
        }
        for (auto i : queries) {
            set<string> s;
            double num = myfind(i.first, i.second, node, s);
            if (num)    ans.push_back(num);
            else    ans.push_back(-1.0);
        }
        return ans;
    }

    double myfind(string first, string second, map<string, map<string,double>> &node, set<string> &s) {
        if (node[first].find(second) != node[first].end())   return node[first][second];
        for (auto i : node[first]) {
            if (s.find(i.first) == s.end()) {
                s.insert(i.first);
                double num = myfind(i.first, second, node, s);
                if (num)    return i.second*num;
            }
        }
        return 0;
    }
};