Evaluate Division

最新推荐文章于 2021-08-25 10:10:21 发布

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本文介绍了一种利用图算法解决变量间关系及查询的方法。通过建立图模型，每个分数被视为一个节点，边权代表乘数值。文章提供了一个具体的实现方案，并附带示例代码，展示了如何求解给定等式中的未知变量关系。

这周是做Graph相关的问题

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"] ].

这个题的主要思想是将每个分数当做图中的一个节点，边权为乘数的权值。基础节点共有2n个，分别是他本身和其倒数。
可以通过基础节点*基础分数或者新增节点*基础分数得到一个新增节点。

代码如下

public:  
    vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values,

                                 vector<pair<string, string>> queries) {   
        unordered_map<string, Node*> map;  
        vector<double> res;  
        for (int i = 0; i < equations.size(); i++) {  
            string s1 = equations[i].first, s2 = equations[i].second;  
            if (map.count(s1) == 0 && map.count(s2) == 0) {  
                map[s1] = new Node();  
                map[s2] = new Node();  
                map[s1] -> value = values[i];  
                map[s2] -> value = 1;  
                map[s1] -> parent = map[s2];  
            } else if (map.count(s1) == 0) {  
                map[s1] = new Node();  
                map[s1] -> value = map[s2] -> value * values[i];  
                map[s1] -> parent = map[s2];  
            } else if (map.count(s2) == 0) {  
                map[s2] = new Node();  
                map[s2] -> value = map[s1] -> value / values[i];  
                map[s2] -> parent = map[s1];  
            } else {  
                unionNodes(map[s1], map[s2], values[i], map);  
            }  
        }  
  
        for (auto query : queries) {  
            if (map.count(query.first) == 0 || map.count(query.second) == 0 || findParent(map[query.first]) !=  
                findParent(map[query.second]))  
                res.push_back(-1);  
            else  
                res.push_back(map[query.first] -> value / map[query.second] -> value);  
        }  
        return res;  
    }  
      
private:  
    struct Node {  
        Node* parent;  
        double value = 0.0;  
        Node()  {parent = this;}  
    };  //用来遍历的节点
      
    void unionNodes(Node* node1, Node* node2, double num, unordered_map<string, Node*>& map) {  
        Node* parent1 = findParent(node1), *parent2 = findParent(node2);  
        double ratio = node2 -> value * num / node1 -> value;  
        for (auto it = map.begin(); it != map.end(); it++) { 
            if (findParent(it -> second) == parent1) {  
                it -> second -> value *= ratio;  
            }  
        }  
        parent1 -> parent = parent2;  
    }  
      
    Node* findParent(Node* node) {  
        if (node -> parent == node)  
            return node;  
        node -> parent = findParent(node -> parent);  
        return node -> parent;  
    }  
};