1466. Reorder Routes to Make All Paths Lead to the City Zero

https://leetcode.com/problems/reorder-routes-to-make-all-paths-lead-to-the-city-zero/

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]

树形有向图问最少翻转多少条边后能让所有结点都能走到结点0。由于是树形，每个结点只能通过离0近的边走到0而不能绕路环走到0，因此从0开始做DFS/BFS，当遇到的边是反着的时候翻转它。

class Solution:
    def minReorder(self, n: int, connections: List[List[int]]) -> int:
        g, g2 = collections.defaultdict(list), collections.defaultdict(list)
        for e in connections:
            g2[e[0]].append(e[1])
            g[e[1]].append(e[0])
        q = collections.deque()
        v = set()
        v.add(0)
        q.append(0)
        res = 0
        while len(q) > 0:
            t = q.popleft()
            for i in g[t]:
                if i in v:
                    continue
                q.append(i)
                v.add(i)
            for i in g2[t]:
                if i in v:
                    continue
                q.append(i)
                v.add(i)
                res += 1
        return res