题目如下:

Consider a directed graph, with nodes labelled 0, 1, ..., n-1.  In this graph, each edge is either red or blue, and there could be self-edges or parallel edges.

Each [i, j] in red_edges denotes a red directed edge from node i to node j.  Similarly, each [i, j] in blue_edges denotes a blue directed edge from node i to node j.

Return an array answer of length n, where each answer[X] is the length of the shortest path from node 0 to node X such that the edge colors alternate along the path (or -1 if such a path doesn't exist).

Example 1:

Input: n = 3, red_edges = [[0,1],[1,2]], blue_edges = []
Output: [0,1,-1]

Example 2:

Input: n = 3, red_edges = [[0,1]], blue_edges = [[2,1]]
Output: [0,1,-1]

Example 3:

Input: n = 3, red_edges = [[1,0]], blue_edges = [[2,1]]
Output: [0,-1,-1]

Example 4:

Input: n = 3, red_edges = [[0,1]], blue_edges = [[1,2]]
Output: [0,1,2]

Example 5:

Input: n = 3, red_edges = [[0,1],[0,2]], blue_edges = [[1,0]]
Output: [0,1,1]

Constraints:

  • 1 <= n <= 100
  • red_edges.length <= 400
  • blue_edges.length <= 400
  • red_edges[i].length == blue_edges[i].length == 2
  • 0 <= red_edges[i][j], blue_edges[i][j] < n

解题思路:本题采用BFS的思想。对于每一个节点来说,分别求出其红边和蓝边作为入口的最小值。

代码如下:

class Solution(object):
def shortestAlternatingPaths(self, n, red_edges, blue_edges):
"""
:type n: int
:type red_edges: List[List[int]]
:type blue_edges: List[List[int]]
:rtype: List[int]
"""
res = [0] + [float('inf')] * (n - 1)
queue = []
red_used = [0] * len(red_edges)
blue_used = [0] * len(blue_edges)
def process(target, edges, res, color,used_list,step_count):
for inx,(i, j) in enumerate(edges):
used = used_list[inx]
if i == target and used == 0:
res[j] = min(res[j],step_count + 1)
queue.append((j, color,step_count + 1))
used_list[inx] = 1
#red
process(0, red_edges, res, 'R',red_used,0)
while len(queue) > 0:
num, color,step = queue.pop(0)
if color == 'R':
process(num, blue_edges, res, 'B',blue_used,step)
else:
process(num, red_edges, res, 'R',red_used,step) red_used = [0] * len(red_edges)
blue_used = [0] * len(blue_edges)
process(0, blue_edges, res, 'B', blue_used,0)
while len(queue) > 0:
num, color,step = queue.pop(0)
if color == 'R':
process(num, blue_edges, res, 'B',blue_used,step)
else:
process(num, red_edges, res, 'R',red_used,step) res = map(lambda x: x if x != float('inf') else -1, res)
return res

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