【leetcode】1129. Shortest Path with Alternating Colors

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
        

标签：Alternating,1129,used,res,blue,step,Colors,edges,red
来源： https://www.cnblogs.com/seyjs/p/11302132.html