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T 秒后青蛙的位置（DFS）

链接

class Solution:
    def frogPosition(self, n: int, edges: List[List[int]], t: int, target: int) -> float:
        g = [[] for _ in range(n + 1)]
        for x, y in edges:
            g[x].append(y)
            g[y].append(x)
        g[1].append(0)
        ans = 0
        def dfs(x, fa, time, prod):
            if x == target and (time == 0 or len(g[x]) == 1):
                nonlocal ans
                ans = 1 / prod
                return True
            if x == target or time == 0: return False
            for y in g[x]:
                if y == fa: continue
                if dfs(y, x, time - 1, prod * (len(g[x]) - 1)): return True
            return False
        dfs(1, 0, t, 1)
        return ans

树上最大得分和路径（DFS）

链接

class Solution:
    def mostProfitablePath(self, edges: List[List[int]], bob: int, amount: List[int]) -> int:
        n = len(amount)
        g = [[] for _ in range(n)]
        for x, y in edges:
            g[x].append(y)
            g[y].append(x)
        g[0].append(-1)
        bob_time = [n] * n
        def dfs_bob(x: int, fa: int, t: int) -> bool:
            if x == 0:
                bob_time[x] = t
                return True
            for y in g[x]:
                if y != fa and dfs_bob(y, x, t + 1):
                    bob_time[x] = t
                    return True
            return False
        dfs_bob(bob, -1, 0)

        ans = -inf
        def dfs_alice(x: int, fa: int, alice_time: int, tot: int) -> None:
            if alice_time < bob_time[x]:
                tot += amount[x]
            elif alice_time == bob_time[x]:
                tot += amount[x] // 2
            if len(g[x]) == 1:
                nonlocal ans
                ans = max(ans, tot)
                return
            for y in g[x]:
                if y != fa:
                    dfs_alice(y, x, alice_time + 1, tot)
        dfs_alice(0, -1, 0, 0)
        return ans