对于树上任意一点来讲，相距最远的点要么是从父节点走过来的，要么就是从儿子节点中走过来。因此可以先进行一次自下向上的dp，处理出从儿子节点过来的最远点的距离。然后再进行一次自上而下的dp，同时传入从父亲节点过来时最远点的距离，这样和从儿子节点过来的最远点的距离对比一下，就可以找到距这个点最远的点的距离了。
INF = 0x3f3f3f3f N = 0 g = [[]] dp = [] ans = [] res = 0 def input(): global N, g N = int(raw_input()) g = [[] for i in range(N + 1)] for i in range(2, N + 1): x = int(raw_input()) g[x].append(i), g[i].append(x) def bottom_up(x, px): global dp, g dp[x] = 0 for y in g[x]: if y == px: continue bottom_up(y, x) dp[x] = max(dp[x], dp[y] + 1) def top_down(x, px, dis): global dp, res, ans, g t = max(dp[x], dis) if t < res: res = t ans = [x] elif t == res: ans.append(x) opt = [dp[i] + 1 for i in g[x]] for i in range(len(opt) - 2, -1, -1): opt[i] = max(opt[i], opt[i + 1]) i = 0 t = dis for y in g[x]: i += 1 if y == px: continue ndis = t if i < len(opt) and opt[i] > ndis: ndis = opt[i] top_down(y, x, ndis + 1) t = max(t, dp[y] + 1) def process(): global N, dp, res, ans dp = [-1 for i in range(N + 1)] bottom_up(1, -1) res = 0x3f3f3f3f top_down(1, -1, 0) ans.sort() for i in ans: print i, print #main input() process()
