title
给你一个变量对数组 equations 和一个实数值数组 values 作为已知条件，其中 equations[i] = [Ai, Bi] 和 values[i] 共同表示等式 Ai / Bi = values[i] 。每个 Ai 或 Bi 是一个表示单个变量的字符串。
另有一些以数组 queries 表示的问题，其中 queries[j] = [Cj, Dj] 表示第 j 个问题，请你根据已知条件找出 Cj / Dj = ? 的结果作为答案。
返回 所有问题的答案 。如果存在某个无法确定的答案，则用 -1.0 替代这个答案。如果问题中出现了给定的已知条件中没有出现的字符串，也需要用 -1.0 替代这个答案。
注意：输入总是有效的。你可以假设除法运算中不会出现除数为 0 的情况，且不存在任何矛盾的结果。
示例 1：
输入：equations = [["a","b"],["b","c"]], values = [2.0,3.0], queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]
输出：[6.00000,0.50000,-1.00000,1.00000,-1.00000]
解释：
条件：a / b = 2.0, b / c = 3.0
问题：a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ?
结果：[6.0, 0.5, -1.0, 1.0, -1.0 ]
示例 2：
输入：equations = [["a","b"],["b","c"],["bc","cd"]], values = [1.5,2.5,5.0], queries = [["a","c"],["c","b"],["bc","cd"],["cd","bc"]]
输出：[3.75000,0.40000,5.00000,0.20000]
示例 3：
输入：equations = [["a","b"]], values = [0.5], queries = [["a","b"],["b","a"],["a","c"],["x","y"]]
输出：[0.50000,2.00000,-1.00000,-1.00000]
提示：
1 <= equations.length <= 20equations[i].length == 21 <= Ai.length, Bi.length <= 5values.length == equations.length0.0 < values[i] <= 20.01 <= queries.length <= 20queries[i].length == 21 <= Cj.length, Dj.length <= 5Ai, Bi, Cj, Dj由小写英文字母与数字组成
solution
# bfs
class Solution:
def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
graph = collections.defaultdict(lambda: [])
n = len(values)
for i in range(n):
a, b = equations[i]
v = values[i]
graph[a].append((b, v))
graph[b].append((a, 1 / v))
def bfs(start, end):
if start not in graph or end not in graph:
return -1
if start == end:
return 1
q = collections.deque()
q.append((start, 1))
visit = set()
visit.add(start)
while len(q) > 0:
top, ans = q.popleft()
for next, v in graph[top]:
if next in visit:
continue
visit.add(next)
if next == end:
return ans * v
q.append((next, ans * v))
return -1
res = []
for a, b in queries:
t = bfs(a, b)
res.append(t)
return res