Gym - 100676H Capital City（边强连通分量 + 树的直径）

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Gym - 100676H Capital City（边强连通分量 + 树的直径）

H. Capital City[ Color: Black ]Bahosain has become the president of Byteland, he is doing his best to make people's liveseasier. Now, he is working on improving road networks between the cities.If two cities are strongly connected, people can use BFS…

CodeForcesGym 100676H Capital City

H. Capital City Time Limit: 3000ms Memory Limit: 262144KB This problem will be judged on CodeForcesGym. Original ID: 100676H64-bit integer IO format: %I64d Java class name: (Any) Bahosain has become the president of Byteland, he is doing his b…

ACM Arabella Collegiate Programming Contest 2015 H. Capital City 边连通分量

题目链接:http://codeforces.com/gym/100676/attachments 题意: 有 n 个点,m 条边,图中,边强连通分量之间可以直达,即距离为 0 ,找一个点当做首都,其他点到首都的最大距离最小. 参考:http://www.cnblogs.com/ost-xg/p/6395100.html 仔细研究了大佬的思路,很牛逼,但是也有一点细节遗漏了,但是还是能AC,interesting: 分析: 边双连通:条件比点双连通松一点,体现在程序中,就是存在一个子节点能到达的…

codeforces GYM 100114 J. Computer Network tarjan 树的直径缩点

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…

Gym - 100781A Adjoin the Networks （树的直径）

题意: n个点,m条边,m <= n <= 100000,边的长度都为1. 点从 0 ~ n-1 编号.开始时图是不连通的,并且没有环. 通过加入一些边后,可以使图连通.要求加入的边不能多余(即生成的图是一棵树). 问连通后的图,任意两点之间的距离的最大值,最小可以是多少? 既然刚开始图不连通也无环,那么就是一些树(特殊情况是点). 于是题目就变成了,如何把很多棵树连起来,使最后生成的树直径最小. 可以想到,如果把两棵直径为 a 和 b 的树加一条边连成一棵,那么直径最小的新树直径为 (a+1…

Gym - 100676H H. Capital City （边双连通分量缩点+树的直径）

https://vjudge.net/problem/Gym-100676H 题意: 给出一个n个城市,城市之间有距离为w的边,现在要选一个中心城市,使得该城市到其余城市的最大距离最短.如果有一些城市是强连通的,那么他们可以使用传送门瞬间到达. 思路:因为强连通时可以瞬移,因为是无向图,所以计算边双连通分量然后重新建图,这样,也就不存在环了. 接下来,计算一下树的直径,因为中心城市肯定选在树的直径上,这样才有可能使最大的边最短. #include<iostream> #include<a…

codeforces GYM 100114 J. Computer Network 无相图缩点+树的直径

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…