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.

给出一些分式以及结果，求若干个分式的值。可以发现，一个分式，比如a/d，可以等于a/b*b/c*c/d，求一个分式的值可以相当于寻找路径。构建一个图，节点的键值为分子或者分母的符号，节点之间的边的值为分式的值。求分式x/y的值时，相当于找出x到y的一条路径，则该路径上全部值的积就是答案。

代码：

class Solution {
public:
vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {
map<string, map<string, double>> m;
for(int i = 0; i < values.size(); ++i) {
m[equations[i].first][equations[i].second] = values[i];
m[equations[i].second][equations[i].first] = 1 / values[i];
}

vector<double> ans;
for(auto query:queries) {
string src = query.first, dst = query.second;
if(m.find(src) == m.end() || m.find(dst) == m.end()) {
ans.push_back(-1.0);
continue;
}
if(src == dst) {
ans.push_back(1.0);
continue;
}
double val = -1;
queue<pair<string, double>> q;
map<string, bool> isvisited;
q.push({src, 1});
while(!q.empty()) {
auto cur = q.front();
q.pop();
if(m.find(cur.first) == m.end()) continue;
if(cur.first == dst) {
val = cur.second;
break;
}
isvisited[cur.first] = true;
for(auto nxt:m[cur.first]) {
if(isvisited[nxt.first]) continue;
q.push({nxt.first, cur.second*nxt.second});
}
}
ans.push_back(val);
}
return ans;
}
};