Codeforces Round #527 (Div. 3) F. Tree with Maximum Cost 【DFS换根 || 树形dp】 a tree consisting exactly of nn vertices. Tree is a connected undirected graph with n−1n−1 edges. Each vertex vv of this tree has a value avav assigned to it. Let dist(x,y)dist(x,y)...
CodeForces - 1092F Tree with Maximum Cost 题意:给出一棵树,每个结点有一个权值。定义一棵树以ai为根节点的价值为 剩下每个结点到根节点的距离乘权值 之和。求这棵树的最大价值。 解:随便选一个结点为根,从下到上统计当前价值,设son[i]为以i结点为子树根节点,这颗子树价值乘路径长度之和。考虑换根,...
首先valuevvaluev相比valueuvalueu,根(v或u)与以v为根的子树中的每一个节点的距离都小了1 在value上表现为 res -= sum[v] 其次在以v为根的子树之外的节点,跟到那些节点的距离都大了1 所以sum[u] -= sum[v], res += sum[u] 此时因为v要成为整个树的根,所以sum[v]+=sum[u] 代码: #include<i...
Codeforces Round #527 (Div. 3) F. Tree with Maximum Cost 【DFS换根 || 树形dp】,程序员大本营,技术文章内容聚合第一站。
[v]++; } for (int i = 1; i <= n; i++) { if (du[i] < 2) { root = i; break; } } dfs1(root, 0); dfs2(root, 0, 0); dfs3(root, 0); int ans = *max_element(dp + 1, dp + 1 + n); cout << ans << endl; } signed main() { ios::sync_with_stdio(false...
CF:Tree with Maximum Cost http://codeforces.com/contest/1092/problem/F 题目: 有一棵树,树的每个节点都有其权值 问一个树上以某个点为树的根节点,其他点到根节点的距离乘上权值最大是多少 思路: 可以运用树状dp解决: 假设suma[x]代表x的子树的权值大小,dp[x]代表x的子树到x的距离乘上权值的大小,...
codeforces 1092——F:Tree with Maximum Cost (树上DP) Tree with Maximum Cost (树上DP) 题目链接 题意: 给定一颗无根树,每点都有权值a[i],定义两点之间的距离dis(x,y)是两点之间简单路径上的边数,求树的最大成本。其中树的成本定义为:假设树的根节点为v,成本为...
CF1092F Tree with Maximum Cost(换根dp) 解题思路 先预处理出以1为根节点的各节点的size。 定义一个节点的size为以这个节点为根的子树的a的和。 然后求出ans[1],再进行换根dp求出每个节点的ans,比较大小输出即可。 AC代码 #include<iostream> #include<cstdio>...
1 query for ith element "kills" all previous type 2 operations for this element, so we only should find a maximum type 2 operation x which is equal to suffix maximum of array of type 2 queries from S[i] and if it is greater than ith element update this element with found maximum ...
You are given a tree with NN (1≤N≤1051≤N≤105) vertices. You need to partition the vertices into minimum number of sets, such that every set has a maximum size of SS (1≤S≤N1≤S≤N) and maximum distance between any of the two vertices, that are in the same set is less ...