Hi All. Is there anyLINEARalgorithm to find the diameter of a graph, which have only one cycle(N vertexes, N paths)? How do you define the diameter? Maximum shortest path between two vertices or simply the longest simple path in the graph? For both ways you divide the problem into two...
但由于理解错一般无向图直径的定义,不会处理介于菊花图与完全图之间的情况。此外,对于完全图直径为1和菊花图直径为2的性质理解的也不够深刻。 改进:对于题目定义的新概念要思考,同时要注意积累特殊情况的性质(异或和、菊花图等)。 #include<cstdio>#include<iostream>#definell long longusingnamespacestd;intmain...
Hi, Codeforces. Could anybody kindly tell me something about fast calculating radius and/or diameter of non-weighted undirected graph (definitions can be foundhere) ? Fast = faster, than in O(MN) (breadth-first searches from each vertex). Any results are welcome: probabilistic, good in averag...
Given a sequence ofnn integersa1,a2,…,ana1,a2,…,an construct aconnected undirected graph ofnn vertices such that: the graph contains no self-loops and no multiple edges; the degreedidi of theii-th vertex doesn't exceedaiai (i.e.di≤aidi≤ai); the diameter of the graph is maximum ...
Codeforces 1082D Maximum Diameter Graph (贪心构造) 题目链接> 题目大意: 给你一些点的最大度数,让你构造一张图,使得该图的直径最长,输出对应直径以及所有的边。 解题分析: 一道比较暴力的构造题,首先,我们贪心的想,要使图的直径最长,肯定是尽可能的将所有的顶点连在同一条链上,并且,所有度数为1的点都只能...
http://codeforces.com/contest/1082/problem/D 先判度数列之和是否大于等于2*n-2 小于则构不成图 然后就是构造直径最长的树 度数大于1的当树干 先连起来 度数为1的只能当叶子 一个一个补到树干上即可 我tm智障啊 加叶节点时忘记先加到端点上了 很烦 ...
It is guaranteed that the given graph is a forest. Each of the nextqlines contains two integersuiandvi(1 ≤ ui, vi ≤ n) — the vertices given in thei-th query. Output For each query print the expected value ofdas described in the problem statement. ...
如果一个点为中心构成一个树,那么当m为n-1时,此时正好每个点的最大距离为2,当m为n*(n+1)/2时,此时正好每个点的最大距离为1,然后只要判断是否能构成图,是否存在重边的情况 #include<bits/stdc++.h> using namespace std; long long t , n , m , k ; int main(){ ...
Codeforces 1082D Maximum Diameter Graph (贪心构造) <题目链接> 题目大意: 给你一些点的最大度数,让你构造一张图,使得该图的直径最长,输出对应直径以及所有的边。 解题分析: 一道比较暴力的构造题,首先,我们贪心的想,要使图的直径最长,肯定是尽可能的将所有的顶点连在同一条链上,并且,所有度数为1的点都只能...
CodeForces 1082 D Maximum Diameter Graph 题目传送门 题意:现在有n个点,每个点的度数最大为di,现在要求你构成一棵树,求直径最长。 题解:把所有度数为2的点先扣出来,这些就是这颗树的主干,也就是最长的距离。 然后我们把度数为2的点连起来,之后就处理1的点,先在主干的最左边和最右边加上新的点,这样可以...