leetcode1466. Reorder Routes to Make All Paths Lead to the City Zero

Posted 2020-12-12 seyjs

题目如下：

There are n cities numbered from 0 to n-1 and n-1 roads such that there is only one way to travel between two different cities (this network form a tree). Last year, The ministry of transport decided to orient the roads in one direction because they are too narrow.

Roads are represented by connections where connections[i] = [a, b] represents a road from city a to b.

This year, there will be a big event in the capital (city 0), and many people want to travel to this city.

Your task consists of reorienting some roads such that each city can visit the city 0. Return the minimum number of edges changed.

It‘s guaranteed that each city can reach the city 0 after reorder.

Example 1:

Input: n = 6, connections = [[0,1],[1,3],[2,3],[4,0],[4,5]]
Output: 3
Explanation: Change the direction of edges show in red such that each node can reach the node 0 (capital).

Example 2:

Input: n = 5, connections = [[1,0],[1,2],[3,2],[3,4]]
Output: 2
Explanation: Change the direction of edges show in red such that each node can reach the node 0 (capital).

Example 3:

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

Constraints:

• 2 <= n <= 5 * 10^4
• connections.length == n-1
• connections[i].length == 2
• 0 <= connections[i][0], connections[i][1] <= n-1
• connections[i][0] != connections[i][1]

解题思路：参考BFS的思想。从起点0开始，遇到[x,0]的connection，则表示需要翻转；如果是[0,x]的格式，则不需要翻转，同时把x加入可到达的集合。然后继续计算与x为起点或者终点的connection。

代码如下：

class Solution(object):
    def minReorder(self, n, connections):
        """
        :type n: int
        :type connections: List[List[int]]
        :rtype: int
        """
        dic = {}
        dic[0] = 1
        res = 0
        while len(connections) > 0:
            start,end = connections.pop(0)
            if start in dic:
                res += 1
                dic[end] = 1
            elif end in dic:
                dic[start] = 1
            else:
                connections.append([start,end])
        return res