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  • 【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
            
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  • 原文地址:https://www.cnblogs.com/seyjs/p/11302132.html
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