Annoying problem 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5296 Description Coco has a tree, whose nodes are conveniently labeled by 1,2,-,n, which has n-1 edge,each edge has a weight. An existing set S is initially empty. Now there are two kin…

Annoying problem Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 203    Accepted Submission(s): 60 Problem Description Coco has a tree, whose nodes are conveniently labeled by 1,2,-,n, which h…

题解链接 Annoying problem Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 480    Accepted Submission(s): 146 Problem Description Coco has a tree, whose nodes are conveniently labeled by 1,2,-,n, w…

Annoying problem Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1006    Accepted Submission(s): 330 Problem Description Coco has a tree, whose nodes are conveniently labeled by 1,2,…,n, which…

Annoying problem Problem Description Coco has a tree, whose nodes are conveniently labeled by 1,2,…,n, which has n-1 edge,each edge has a weight. An existing set S is initially empty.Now there are two kinds of operation: 1 x: If the node x is not in…

题意: 给一棵n个节点的树,再给q个操作,初始集合S为空,每个操作要在一个集合S中删除或增加某些点,输出每次操作后:要使得集合中任意两点互可达所耗最小需要多少权值.(记住只能利用原来给的树边.给的树边已经有向.10万个点,10万个操作) 思路:只能用 O(nlogn)的复杂度.官方题解: 重点也就是要找到集合S中的以x和y为端点一条链,使得操作点u到达这条链是最近的.删除也是这样,找到这条链,删除u到这链的路长. 步骤: (1)记录从根遍历的DFS序. (2)计算每个点到根的路径所经过边的权之和…

Annoying problem 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5293 Description Coco has a tree, whose vertices are conveniently labeled by 1,2,-,n. There are m chain on the tree, Each chain has a certain weight. Coco would like to pick out some ch…

dfs一遍得到每一个节点的dfs序,对于要插入的节点x分两种情况考虑: 1,假设x能够在集合中的某些点之间,找到左边和右边距离x近期的两个点,即DFS序小于x的DFS序最大点,和大于x的DFS序最小的点...... 2.假设x在集合中的点某一側,则找距离x的dfs序最小和最大的点 将x插入这个集合最少要走的距离为 dist[x]-dist[LCA(left,x)]-dist[LCA(right,x)]+dist[LCA(left,right)] 删除同理 Annoying problem Tim…

LCA+RMQ.挺不错的一道题目. 思路是如何通过LCA维护费用.当加入新的点u是,费用增量为dis[u]-dis[lca(u, lower_u)] - dis[lca(u, greater_u)] + dis[lca(lower_u, greater_u)].若beg[u]大于当前最大值或小于最小值,lower_u=min of current, greater_u = max of current. /* 5296 */ #include <iostream> #include <s…

Annoying problem Problem's Link: http://acm.hdu.edu.cn/showproblem.php?pid=5296 Mean: 给你一个有根树和一个节点集合S,初始S为空,有q个操作,每个操作要么从树中选一个结点加入到S中(不删除树中节点),要么从S集合中删除一个结点.你需要从树中选一些边组成集合E,E中的边能够是S中的节点连通.对于每一个操作,输出操作后E中所有边的边权之和. analyse: 首先是构图,把树看作一个无向图,使用邻接表存图. 处理出…