Algorithm-week4

Week4

Program--Medium--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.

题目解析：

这是一个非常明显的图算法的题目。题目给出了点到点的路径以及路径的权值，对于一个查询由起点到终点，找到一条路径由起点去往终点，并计算路径上的权值的积。解法也很直接：

1.首先我们得将图的点以及权值以方便我们访问的方式存储下来，类似课上的路径集合E。在这里，我用一个vector<string, vector<Node>>来储存每个点能到的点的集合，Node上有点的名字以及到它的权值，这样就可以非常方便地访问边以及对应的权值。

2.我们建好了图的结构信息，下一步就是如何查找路径以及返回权值的积。在这里我用的方法是BFS，对每个点进行一次遍历，用vector<string, pair<string, double>>来存储遍历的当前节点的前一节点名以及路径的权值。

3.最后一步用遍历所得的Pre信息来从终点进行反向搜索，不断返回遍历时的上一个几点，直到找到起始点或者没有路径可走为止。

4.如果成功回到起点，证明存在路径从起点到终点，否则不存在路径则返回-1，特殊情况是起点等于终点则直接输出1，其中一个点不存在直接输出-1。

代码：

class Solution {public:    struct Node{        string name;        double data;        Node(string n, double d) {            name = n;            data = d;        }    };    map<string, vector<Node>> Point;    map<string, pair<string, double>> Pre;    vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {        for (int i = 0; i < equations.size(); i++) {            pair<string, string> temp = equations[i];            double tempValue = values[i];            Node tempNode(temp.second, tempValue);            Node tempNode2(temp.first, 1/tempValue);            Point[temp.first].push_back(tempNode);            Point[temp.second].push_back(tempNode2);        }        vector<double> result;        cout << "success" << endl;        for (int i = 0; i < queries.size(); i++) {            string end;            queue<string> source;            Pre[queries[i].first].first = queries[i].first;            Pre[queries[i].second].second = 1;            source.push(queries[i].first), end = queries[i].second;            //cout << "i = " << i << endl;                        if (!Point[queries[i].first].size() || !Point[end].size()) {                result.push_back(-1.0);            }            else {                result.push_back(Query(source, end));            }        }                return result;    }        double Query(queue<string> s, string e) {        string start = s.front();        while(!s.empty()) {            string temp = s.front();            s.pop();            cout << s.size() << endl;            for (int i = 0; i < Point[temp].size(); i++) {                if (Pre[temp].first != Point[temp][i].name) {                    s.push(Point[temp][i].name);                    Pre[Point[temp][i].name].first = temp;                    Pre[Point[temp][i].name].second = Point[temp][i].data;                }            }        }        string temp = e;        string last = e;        double num = 1;        while (Pre[temp].first != temp && Pre[temp].first != ""){            temp = Pre[last].first;            num *= Pre[last].second;            //cout << "last = " << last << " temp = " << temp << endl;            last = temp;            //cout << "num = " << num << endl;        }        if (temp != start) {            return -1.0;        }        else {            return num;        }    }    };

完成度：