2477. Minimum Fuel Cost to Report to the

[개발뉴비(coderhs)]

class Solution:

    def dfs(self, cur, visit, dp, graph):

        for next in graph[cur]:

            if visit[next] == False:

                visit[next] = True

                dp[cur] += self.dfs(next, visit, dp, graph)

        return dp[cur]

    def minimumFuelCost(self, roads: List[List[int]], seats: int) -> int:

        n = len(roads) + 1

        graph = [[] for _ in range(n)]

        visit = [False for _ in range(n)]

        dp = [1 for _ in range(n)]

        for l, r in roads:

            graph[l].append(r)

            graph[r].append(l)

        visit[0] = True

        self.dfs(0, visit, dp, graph)

        return sum([(x + seats - 1) // seats for x in dp[1:]])

열심히 엄청 길게 풀었는데 알고 보니 내가 문제를 잘못 이해했었음

다시 짜야했다 ㅜㅜ 그냥 평범한 트리 dp였네