标签：get eva orm ons uil NPU real rap solution
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.
1 class Solution { 2 public: 3 vector<double> calcEquation(vector<pair<string, string>> &equations, vector<double>& values, vector<pair<string, string>> &queries) { 4 unordered_map<string,int>vertices = getVertices(equations); 5 vector<vector<double>> graph = buildGraph(vertices, equations, values); 6 7 vector<double> result(queries.size()); 8 vector<bool> visited(vertices.size()); 9 for(int i=0;i<queries.size();i++){ 10 string from = queries[i].first; 11 string to = queries[i].second; 12 if(vertices.find(from)==vertices.end()||vertices.find(to)==vertices.end()){ 13 result[i]=-1; 14 }else{ 15 fill(visited.begin(),visited.end(),false); 16 result[i]=dfs(vertices[from], vertices[to], graph, visited); 17 } 18 } 19 20 return result; 21 } 22 private: 23 //Mapping from string to integer 24 unordered_map<string,int> getVertices(vector<pair<string,string>>& equations){ 25 unordered_map<string,int> map; 26 int i = 0; 27 for(auto equation: equations){ 28 if(map.find(equation.first)==map.end()){ 29 map[equation.first] = i; 30 i++; 31 } 32 if(map.find(equation.second)==map.end()){ 33 map[equation.second] = i; 34 i++; 35 } 36 } 37 return map; 38 } 39 40 vector<vector<double>> buildGraph(unordered_map<string,int>& vertices, vector<pair<string,string>>& equations, vector<double>& values){ 41 vector<vector<double>> graph (vertices.size(),vector<double>(vertices.size())); 42 for(int i=0;i<vertices.size();i++){ 43 graph[i][i] = 1.0; 44 } 45 for(int i=0;i<equations.size();i++){ 46 int a = vertices[equations[i].first]; 47 int b = vertices[equations[i].second]; 48 graph[a][b] = values[i]; 49 graph[b][a] = 1.0/values[i]; 50 } 51 52 return graph; 53 } 54 55 double dfs(int from, int to, vector<vector<double>>& graph, vector<bool>& visited){ 56 for(int j=0; j<graph[from].size();j++){ 57 if(graph[from][j]!=0&&!visited[j]){ 58 if(j==to){ 59 return graph[from][to]; 60 }else{ 61 visited[j] = true; 62 double result = dfs(j,to,graph,visited); 63 if(result!=-1){ 64 return graph[from][j]*result; 65 } 66 visited[j]=false; 67 } 68 } 69 } 70 71 return -1; 72 } 73 };
标签：get eva orm ons uil NPU real rap solution
原文地址：https://www.cnblogs.com/ruisha/p/9222504.html
