動機
建圖去走
Problem
You are given an array of variable pairs equations
and an array of real numbers values
, where equations[i] = [Ai, Bi]
and values[i]
represent the equation Ai / Bi = values[i]
. Each Ai
or Bi
is a string that represents a single variable.
You are also given some queries
, where queries[j] = [Cj, Dj]
represents the jth
query where you must find the answer for Cj / Dj = ?
.
Return the answers to all queries. If a single answer cannot be determined, return -1.0
.
Note: The input is always valid. You may assume that evaluating the queries will not result in division by zero and that there is no contradiction.
Example 1:
Input: equations = [["a","b"],["b","c"]], values = [2.0,3.0], queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]Output: [6.00000,0.50000,-1.00000,1.00000,-1.00000]Explanation: Given: a / b = 2.0, b / c = 3.0queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ?return: [6.0, 0.5, -1.0, 1.0, -1.0 ]
Example 2:
Input: equations = [["a","b"],["b","c"],["bc","cd"]], values = [1.5,2.5,5.0], queries = [["a","c"],["c","b"],["bc","cd"],["cd","bc"]]Output: [3.75000,0.40000,5.00000,0.20000]
Example 3:
Input: equations = [["a","b"]], values = [0.5], queries = [["a","b"],["b","a"],["a","c"],["x","y"]]Output: [0.50000,2.00000,-1.00000,-1.00000]
Constraints:
1 <= equations.length <= 20
equations[i].length == 2
1 <= Ai.length, Bi.length <= 5
values.length == equations.length
0.0 < values[i] <= 20.0
1 <= queries.length <= 20
queries[i].length == 2
1 <= Cj.length, Dj.length <= 5
Ai, Bi, Cj, Dj
consist of lower case English letters and digits.
Sol
class Solution:
def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
db = defaultdict(lambda :-1.0)
gh = defaultdict(list)
for ((a,b),val) in zip(equations, values):
db[(a,b)] = val
db[(b,a)] = 1.0/val
db[(a,a)] = db[(b,b)] = 1.0
gh[a].append(b)
gh[b].append(a)
def dfs(root,lhs, acc=1.0, seen=set()):
if lhs in seen:
return
else:
seen.add(lhs)
for rhs in gh[lhs]:
db[(root,rhs)] = acc*db[(lhs,rhs)]
db[(rhs,root)] = 1/(acc*db[(lhs,rhs)])
dfs(root,rhs,acc*db[(lhs,rhs)],seen)
[dfs(root,root,1.0,set()) for root in gh.keys()]
return [db[(lhs,rhs)] for (lhs,rhs) in queries]