给一个n个节点m条边的树, 每条边有权值, 输出m个数, 每个数代表包含这条边的最小生成树的值。 先将最小生成树求出来, 把树边都标记。 然后对标记的边的两个端点, 我们add(u, v), add(v, u)。 对于每一次输出, 如果这条边被标记了, 那么直接输出mst的值。 否则, 加上这条边之后一定会出现一个...
But, here I want to invite you guys to the contest which begins in an hour. (Sorry for such a short notice!) The contest is based on Minimum Spanning Trees. The link to the contest:Code Monk: Minimum Spanning Tree The setter and the translator of the contest is:shef_2318and the test...
Connected undirected weighted graph without self-loops and multiple edges is given. Graph containsnvertices andmedges. For each edge(u, v) find the minimal possible weight of the spanning tree that contains the edge(u, v). The weight of the spanning tree is the sum of weights of all...
E. Minimum spanning tree for each edge (codeforces.com)codeforces.com/contest/609/problem/E 题目大意 给定一张无向带权图,保证没有重边或者自环。对于每条边(u,v),找出包含这条边的最小生成树的权重。 思路 考虑一条边(u,v),若其本来作为原图的最小生成树中的其中一条边,那么显然对应的答案为原...
n=2^x1+ 2^x2 +2^x3+… 所以对于任意一个整数,都能通过按2的次幂跳来实现。 这里给出超多解析的代码 例题:CF Minimum spanning tree for each edge
First off, build the minimum spanning tree, then, for each edge of the minimum spanning tree, build a new one discarding such edge. The minimum one will be the second minimum spanning tree. Using Kruskal algorithm and a O(n * log(n)) sorting algorithm for the edges, this strategy lead...
2019-12-07 19:35 −Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly two or&nbs... lychnis 0 221 [CF438E]The Child and Binary Tree 2019-12-24 14:44 −[传送门](https://codeforces.com/problemset/probl...
【Codeforces 1086B】Minimum Diameter Tree 我是链接,点我呀:) 【题意】 题意 【题解】 统计叶子节点个数m 把每条和叶子节点相邻的边权设置成s/cnt就可以了 这样答案就是2*s/m(直径最后肯定是从一个叶子节点开始,到另外一个叶子节点结束) 证明:
output standard output Connected undirected weighted graph without self-loops and multiple edges is given. Graph containsnvertices andmedges. For each edge(u, v) find the minimal possible weight of the spanning tree that contains the edge(u, v). ...
题目链接:http://codeforces.com/problemset/problem/609/E 大致就是有一棵树,对于每一条边,询问包含这条边,最小的一个生成树的权值。 做法就是先求一次最小生成树,标记最小生成树上的边,对于这些边,直接就是原始最小生成树。否则必然可以在去掉u到v路径上最长边