graph dfs

Author: idf

399 Evaluate Division.py

@@ -0,0 +1,83 @@
+"""
+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"] ].
+
+"""
+from collections import defaultdict
+from itertools import izip
+
+__author__ = 'Daniel'
+
+
+class Solution(object):
+    def calcEquation(self, equations, values, queries):
+        """
+        transitive closure
+        :type equations: List[List[str]]
+        :type values: List[float]
+        :type queries: List[List[str]]
+        :rtype: List[float]
+        """
+        G = defaultdict(dict)
+        for edge, val in izip(equations, values):
+            s, e = edge
+            G[s][e], G[e][s] = val, 1/val
+            G[s][s], G[e][e] = 1, 1
+
+        return [self.dfs(G, s, e, set()) for s, e in queries]
+
+    def dfs(self, G, s, e, path):
+        if s not in G or e not in G:
+            return -1.0
+        if e in G[s]:
+            return G[s][e]
+        for nbr in G[s]:
+            if nbr not in path:
+                path.add(nbr)
+                val = self.dfs(G, nbr, e, path)
+                if val != -1.0:
+                    return val * G[s][nbr]
+                path.remove(nbr)
+
+        return -1.0
+
+
+class Solution(object):
+    def calcEquation(self, equations, values, queries):
+        """
+        Floyd-Warshall algorithm
+        transitive closure
+        :type equations: List[List[str]]
+        :type values: List[float]
+        :type queries: List[List[str]]
+        :rtype: List[float]
+        """
+        G = defaultdict(dict)
+        for edge, val in izip(equations, values):
+            s, e = edge
+            G[s][e], G[e][s] = val, 1/val
+            G[s][s], G[e][e] = 1, 1
+
+        # Floyd-Warshall
+        for mid in G:
+            for s in G[mid]:
+                for e in G[mid]:
+                    G[s][e] = G[s][mid] * G[mid][e]
+
+        return [G[s].get(e, -1.0) for s, e in queries]
+