5132. Shortest Path with Alternating Colors_consider a directed graph , where each edge is col-优快云博客

本文链接：https://blog.youkuaiyun.com/zjucor/article/details/96705270

本文介绍了一种使用广度优先搜索(BFS)算法求解从节点0到其他各节点的最短路径问题的方法，其中路径上的边颜色需交替出现。通过构建红色与蓝色边的邻接表，并利用双队列实现不同颜色边的交替遍历，最终得到所有可达节点的最短交替路径长度。

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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

思路：BFS

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]
        """
        adj_red={}
        for s,t in red_edges: 
            if s not in adj_red: adj_red[s]=set()
            adj_red[s].add(t)
        adj_blue={}
        for s,t in blue_edges: 
            if s not in adj_blue: adj_blue[s]=set()
            adj_blue[s].add(t)
        
        res={0:0}
        q=[(0,0),(0,1)]
        vis=set(q)
        qq=[]
        step=1
        while q:
            while q:
                s,f=q.pop()
                target=adj_red if f==0 else adj_blue
                for t in target.get(s,[]):
                    if (t,1-f) in vis: continue
                    qq.append((t,1-f))
                    vis.add((t,1-f))
                    if t not in res: res[t]=step
            q,qq=qq,q
            step+=1
        
        return [res[i] if i in res else -1 for i in range(n)]