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题目链接: http://codeforces.com/gym/100114 Description The computer network of “Plunder & Flee Inc.” consists of n servers and m two-way communication links. Two servers can communicate either through a direct link, or through a chain of links, by relayi…
J. Computer Network Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100114 Description The computer network of “Plunder & Flee Inc.” consists of n servers and m two-way communication links. Two servers can communicate either thr…
题目大意:给N个点,求每个点的与其他点距离最大值 很经典的树形dp...很久前就想写来着...看了陈老师的code才会的...mx[x][0], mx[x][1]分别表示x点子树里最长的2个距离, dfs一遍得到. mx[x][2]表示从x的父亲到x的最长路径长度, 也是dfs一遍得到(具体看代码).最后答案就是max(mx[x][0], mx[x][2]). 时间复杂度O(N) ----------------------------------------------------------…
Andrewid the Android is a galaxy-known detective. Now he is preparing a defense against a possible attack by hackers on a major computer network. In this network are n vertices, some pairs of vertices are connected by m undirected channels. It is pla…
https://odzkskevi.qnssl.com/b660f16d70db1969261cd8b11235ec99?v=1537580031 [2012-2013 ACM Central Region of Russia Quarterfinal Programming Contest][J]computer network 题意: n个点,m条边,构成一个无向图,现在让你再任意连接两个点,使得整个图的割边最少. 1 ≤ n ≤ 10 000; 1≤ m ≤ 100 000; 1 ≤ xi…
Computer Network Homework2’s hard question 2. What is the signal which is used to modulate the original signal? A. analog signal B. digital signal C. carrier signal D. base signal (2 points) Answer: C Explanation: original signal指最原始的想要被传给别人的信号 5.In…
Computer Network Homework3’ s hard question 1. Which kind of protocol does CSMA belong to? A. Random Access Protocol B. Take Turns Protocol C. Centralized Protocol (2 points) Answer: A 6. Which provides connection-oriented service? A. LLC1 B. LLC2 C.…
[Codeforces 555E]Case of Computer Network(Tarjan求边-双连通分量+树上差分) 题面 给出一个无向图,以及q条有向路径.问是否存在一种给边定向的方案,使得这q条路径都能被满足.(如果有一条边是从a->b),而经过它的路径是从b->a,那么久不满足).只需要判断,不用输出方案. 分析 对于一个有向环,显然它可以允许各个方向的路径通过.所以我们只要把无向图里的边-双联通分量建成环,然后就不用考虑了.影响答案的只有桥. 所以我们求出所有桥,然后缩点,把图…
Sgu149 Computer Network 题目描述 给你一棵N(N<=10000)个节点的树,求每个点到其他点的最大距离. 不难想到一个节点到其他点的最大距离为:max(以它为根的子树的最大深度,根节点到它的距离+根节点的另外所有的子树(没有这个点的子树)的最大深度). 我们不妨分两次求解出上面的两个答案的元素.显然第一个很好求,我们直接看第二个. 设整棵树的最大深度为dep1,次大深度为dep2,根节点为r.如果当前节点u不在最大深度的那棵子树上,那么它的第二个答案元素的最大距离就为:d…