<题目链接> 题目大意:给你一棵带有边权的树,然后进行q次查询,每次查询输出指定两个节点之间的距离. 解题分析:本题有多重解决方法,首先,可用最短路轻易求解.若只用LCA解决本题,也有三种常用的做法,具体方法如下: LCA转RMQ解法: #include <cstdio> #include <cmath> #include <vector> #include <cstring> #include <algorithm> using n…

Tree Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others) Total Submission(s): 1643    Accepted Submission(s): 461 Problem Description   Zero and One are good friends who always have fun with each other. This time, t…

How far away ? Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description There are n houses in the village and some bidirectional roads connecting them. Every day peole always like…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2586 Problem Description There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2586 题意: 给出一棵 n 个节点的带边权的树, 有 m 个形如 x y 的询问, 要求输出所有 x, y节点之间的最短距离. 思路: dis[i] 存储 i 节点到根节点的最短距离, lca 为 x, y 的最近公共祖先, 那么 x, y 之间的最短距离为: dis[x] + dis[y] - 2 * dis[lca] . 解法1: tarjan离线算法 关于该算法 Tarjan(u)//marg…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2586 题目大意:在一个无向树上,求一条链权和. 解题思路: 0 | 1 /   \ 2      3 设dist[i]为i到根0的链和,求法(Dfs过程中dist[v]=dist[u]+e[i].w) 对于树中任意两点形成的链,可以通过LCA最近公共祖先剖分. 比如2->3,就可以经过LCA点1:  2->1->3 链和=dist[u]+dist[v]-2*dist[LCA[u,v]] (…

题目大意: 多点形成一棵树,树上边有权值,给出一堆询问,求出每个询问中两个点的距离 这里求两个点的距离可以直接理解为求出两个点到根节点的权值之和,再减去2倍的最近公共祖先到根节点的距离 这是自己第一道lca题目 学习了两种方法 第一种在ST算法,利用RMQ思想预处理 /*在线ST算法*/ #pragma comment(linker, "/STACK:102400000,102400000") #include <cstdio> #include <iostream&…

HDU - 2586 How far away ? Time Limit: 1000MS   Memory Limit: 32768KB   64bit IO Format: %I64d & %I64u Submit Status Description There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like th…

hdu 2586 How far away ?倍增LCA 题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=2586 思路: 针对询问次数多的时候,采取倍增求取LCA,同时跟新距离数组 因为 \(2^{16} > 40000\) 所以所以表示祖先的数组dp[][]第二维取到16即可 就这道题来说,与比较tarjan比较,稍快一点 代码: #include <iostream> #include <algorithm> #includ…

HDU 2586 How far away ? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11320    Accepted Submission(s): 4119 Problem Description   There are n houses in the village and some bidirectional roads…