题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2874 题意: 给出 n 个顶点 m 条边的一个森林, 有 k 个形如 x y 的询问, 输出 x, y 之间的最短路径. 思路: 如果将森林换成一棵树的话就是一道 lca 模板题了, 不过本题需要稍作改动. 解法1: tarjan 只需要先判断一下 x, y 是否在一颗树里面就 ok 了, 不过这道题的询问有点多, 很容易 mle. 代码: #include <iostream> #includ…
COT - Count on a tree #tree You are given a tree with N nodes.The tree nodes are numbered from 1 to N.Each node has an integer weight. We will ask you to perform the following operation: u v k : ask for the kth minimum weight on the path from node u …
How far away ? Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 25408 Accepted Submission(s): 10111 Problem Description There are n houses in the village and some bidirectional roads connectin…
Nearest Common Ancestors Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 20715 Accepted: 10910 Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: In the figure, each…
任意门:http://poj.org/problem?id=1986 Distance Queries Time Limit: 2000MS Memory Limit: 30000K Total Submissions: 16648 Accepted: 5817 Case Time Limit: 1000MS Description Farmer John's cows refused to run in his marathon since he chose a path much t…
第一题LCA,代码参考自:Ice_Crazy 思路: 这个最短路算法是想都别想了,可以看出这幅图就是树嘛,那么对于查询就是求树上两个结点最短距离. 这里就是利用LCA的tarjan离线算法. 算法的大致流程: 对于每一点u, ① :建立以u为代表元素的集合. ② :遍历与u相连的结点v,如果没有访问过,对与v使用Tarjan-LCA算法,结束后,将v的集合并入u的集合. ③ :对于与u相关的询问(u,v),如果v被访问过,则结果就是v所在集合的代表. 在这里还需要算距离,需要深度:dis<…
