Ural 1018 Binary Apple Tree (Tree dp | classic)

Url-1018 to give a edge of the value of the Binary Tree, node number is 1 ~ N, 1 is the root node. Cut down some edges and keep only q edges. The root node of the subtree composed of q edges must be 1. What is the maximum weight of this subtree? The idea is a classic tree dp question. According to the questions I have done so far, many of them are derived from this question. F (I, j) indicates subtree I, and retains the maximum weight of j nodes (note that it is a node. The weight of each edge, which is considered as the weight of the son node of the connected two nodes. Then, we can group all the I subtree, that is, each subtree can choose 1, 2,... J-1 side allocated to it. Status transfer: f (I, j) = max {f (I, j-k) + f (v, k) | 1 <= k <j} | v is the son of I} ans = f (1, q + 1) Code

/** =================================================== =================== * This is a solution for ACM/ICPC problem ** @ source: ural-1018 Binary Apple Tree * @ description: Tree dp * @ author: shuangde * @ blog: blog.csdn.net/shuangde800 * @ email: zengshuangde@gmail.com * Copyright (C) 2013/09/01 All rights reserved. * ===================================================== ===================*/# include <iostream >#include <cstdio> # Include <algorithm> # include <vector> # include <cstring> # define MP make_pairusing namespace std; typedef pair <int, int> PII; typedef long int64; const int INF = 0x3f3f3f; const int MAXN = 110; vector <PII> adj [MAXN]; int n, q; int tot [MAXN]; int f [MAXN] [MAXN]; int dfs (int u, int fa) {tot [u] = 1; for (int e = 0; e <adj [u]. size (); ++ e) {int v = adj [u] [e]. first; if (v = fa) continue; tot [u] + = dfs (v, U) ;}for (int e = 0; e <adj [u]. size (); ++ e) {int v = adj [u] [e]. first; int w = adj [u] [e]. second; if (v = fa) continue; for (int I = tot [u]; I> 0; -- I) {for (int j = 1; j <I & j <= tot [v]; ++ j) {f [u] [I] = max (f [u] [I], f [u] [I-j] + f [v] [j] + w) ;}}return tot [u] ;}int main () {while (~ Scanf ("% d", & n, & q) {for (int I = 0; I <= n; ++ I) adj [I]. clear (); for (int I = 0; I <n-1; ++ I) {int u, v, w; scanf ("% d ", & u, & v, & w); adj [u]. push_back (MP (v, w); adj [v]. push_back (MP (u, w);} memset (f, 0, sizeof (f); dfs (1,-1); printf ("% d \ n ", f [1] [q + 1]);} return 0 ;}