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.

写了逻辑错误的代码，下次 别犯这种错误：

import java.util.HashMap;//这个逻辑不是对的  要用搜索的方式public class Solution {  public double[] calcEquation(String[][] equations, double[] values, String[][] queries) {  HashMap<String,Integer> map=new HashMap<String,Integer>();  int index=0;  for(int i=0;i<equations.length;i++){  if(!map.containsKey(equations[i][0])) map.put(equations[i][0],index++);  if(!map.containsKey(equations[i][1])) map.put(equations[i][1],index++);  }  double[][] graph=new double[index][index];  for(int i=0;i<graph.length;i++){  for(int j=0;j<graph[0].length;j++){  if(i!=j)graph[i][j]=-1.0;  else graph[i][j]=1;  }  }  for(int i=0;i<equations.length;i++){  int x=map.get(equations[i][0]);  int y=map.get(equations[i][1]);  graph[x][y]=values[i];  graph[y][x]=1/values[i];  for(int j=0;j<graph[0].length;j++){  if(graph[j][x]!=-1.0) {  graph[j][y]=graph[j][x]*graph[x][y];  graph[y][j]=1/graph[j][y];  }  if(graph[j][y]!=-1.0){  graph[j][x]=graph[j][y]*graph[y][x];  graph[x][j]=1/graph[j][x];  }   }  }  double[] re=new double[queries.length];  for(int i=0;i<queries.length;i++){  Integer x=map.get(queries[i][0]);  Integer y=map.get(queries[i][1]);  if(x==null||y==null) re[i]=-1.0;  else re[i]=graph[x][y];  }  return re;  }}

ac代码：

public class Solution2 {    public double[] calcEquation(String[][] equations, double[] values, String[][] queries) {  HashMap<String,Integer> map=new HashMap<String,Integer>();  int index=0;  for(int i=0;i<equations.length;i++){  if(!map.containsKey(equations[i][0])) map.put(equations[i][0],index++);  if(!map.containsKey(equations[i][1])) map.put(equations[i][1],index++);  }  graph=new double[index][index];  for(int i=0;i<graph.length;i++){  for(int j=0;j<graph[0].length;j++){  if(i!=j)graph[i][j]=-1.0;  else graph[i][j]=1;  }  }  for(int i=0;i<equations.length;i++){  int x=map.get(equations[i][0]);  int y=map.get(equations[i][1]);  graph[x][y]=values[i];  graph[y][x]=1/values[i];  }  double[] re=new double[queries.length];  for(int i=0;i<queries.length;i++){  Integer x=map.get(queries[i][0]);  Integer y=map.get(queries[i][1]);  if(x==null||y==null) re[i]=-1.0;  else{  boolean[] flag=new boolean[graph.length];  flag[x]=true;  re[i]=dfs(x,y,flag);  }  }  return re; } double[][] graph; public double dfs(int s,int t,boolean[] flag){ if(graph[s][t]!=-1.0) return graph[s][t]; double re=-1.0d; for(int i=0;i<graph.length;i++){ if(flag[i]) continue;             flag[i]=true;  if(graph[s][i]!=-1.0){  double temp=dfs(i,t,flag);  if(temp!=-1.0) re=graph[s][i]*temp;  if(re!=-1.0)  return graph[s][t]=re;  }             flag[i]=false; } return graph[s][t]=re; }}