HDU3538 A sample Hamilton path

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HDU3538 A sample Hamilton path

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 503 Accepted Submission(s): 200 Problem Description Give you a Graph,you have to start at the city with ID zero. Input The first line is n(1<…

AGC018D Tree and Hamilton Path(树+树的重心)

题目大意: 给你一棵n个结点树,然后根据这棵树构造一个完全图,求完全图的一条最长的哈密顿路径. 构造方式是,完全图中的dis(u, v)就等于树上的u和v的距离. 题解: 这...这..不就是杜教的那个题还是弱化版的orz 需要注意的是,不是完全一样,这个题求的是哈密顿回路,需要删除一个最小的路径答案就很简单了,找到重心以后. 如果有一棵子树大小为(n+1)/2,那么就只能删除连向那个子树的边如果没有,那么就遍历一遍所有连向重心的边,选一个最小的删除即可. 杜教题连接 http://www…

AtCoder Grand Contest 018 D - Tree and Hamilton Path

题目传送门:https://agc018.contest.atcoder.jp/tasks/agc018_d 题目大意: 给定一棵\(N\)个点的带权树,求最长哈密顿路径(不重不漏经过每个点一次,两点之间转移可以看做瞬移,对答案贡献为两点之间的距离) 这道题直接计算不好算,我们考虑每条边的贡献,基于一种贪心的思想,我们发现只要围着树的重心跑,就可以使每条边得到充分利用考虑边i的贡献,我们假定边i割掉后分成两个大小为x,y的联通块,那么贡献则为\(2*v[i]*min(x,y)\) 因为我们走的…

D - Tree and Hamilton Path

题意给一棵树,问一个排列,使得按顺序走过这些点的路径最长. N<=100000 解法为了能让每条边被经过的次数达到上界, 我们首先找出重心, 然后容易得出一种排列方案,使得答案为以重心为根的深度和*2. 但由于这不是一条回路,还要减去与重心相连的一条边. 如果重心存在一个儿子的size是点数的一半,那么就减去这条儿子的边权. 否则就挑一个最短的边权删掉即可.…

POJ2288 Islands and Bridges

Description Given a map of islands and bridges that connect these islands, a Hamilton path, as we all know, is a path along the bridges such that it visits each island exactly once. On our map, there is also a positive integer value associated with e…

【状压dp】Islands and Bridges

Islands and Bridges Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 11034 Accepted: 2866 Description Given a map of islands and bridges that connect these islands, a Hamilton path, as we all know, is a path along the bridges such that…

[poj2288] Islands and Bridges (状压dp)

Description Given a map of islands and bridges that connect these islands, a Hamilton path, as we all know, is a path along the bridges such that it visits each island exactly once. On our map, there is also a positive integer value associated with e…