給出方程式A / B = k, 其中A 和B 均為用字串表示的變數，k 是一個浮點型數字。根據已知方程式求解問題，並返回計算結果。如果結果不存在，則返回-1.0。

輸入總是有效的。你可以假設除法運算中不會出現除數為 0 的情況，且不存在任何矛盾的結果。

示例 1：

輸入：equations = [["a","b"],["b","c"]], values = [2.0,3.0], queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]
輸出：[6.00000,0.50000,-1.00000,1.00000,-1.00000]
解釋：
給定：a / b = 2.0, b / c = 3.0
問題：a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ?
返回：[6.0, 0.5, -1.0, 1.0, -1.0 ]
示例 2：

輸入：equations = [["a","b"],["b","c"],["bc","cd"]], values = [1.5,2.5,5.0], queries = [["a","c"],["c","b"],["bc","cd"],["cd","bc"]]
輸出：[3.75000,0.40000,5.00000,0.20000]
示例 3：

輸入：equations = [["a","b"]], values = [0.5], queries = [["a","b"],["b","a"],["a","c"],["x","y"]]
輸出：[0.50000,2.00000,-1.00000,-1.00000]

提示：

1 <= equations.length <= 20
equations[i].length == 2
1 <= equations[i][0].length, equations[i][1].length <= 5
values.length ==equations.length
0.0 <values[i] <= 20.0
1 <= queries.length <= 20
queries[i].length == 2
1 <= queries[i][0].length, queries[i][1].length <= 5
equations[i][0], equations[i][1],queries[i][0], queries[i][1] 由小寫英文字母與數字組成

Floyd

from collections import defaultdict

class Solution:
    def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
        edges=defaultdict(dict)
        verts=set()

        for (a,b),c in zip(equations,values):
            verts.add(a)
            verts.add(b)

            edges[a][b]
 
=c
            edges[b][a]=1/c

        for k in verts:
            ek=edges[k]
            for i in verts:
                if i==k:
                    continue
                
                ei=edges[i]
                for j in verts:
                    if k==j or i==j or j in ei:
                        continue

                    if k in ei and j in ek:
                        ei[j]=ei[k]*ek[j]
                
        res=[]
        for a,b in queries:
            if a in edges and b in edges[a]:
                res.append(edges[a][b])
            elif a==b and a in verts and b in verts:
                res.append(1)
            else:
                res.append(-1)

        return res