Cerror is the mayor of city HangZhou. As you may know, the traffic system of this city is so terrible, that there are traffic jams everywhere. Now, Cerror finds out that the main reason of them is the poor design of the roads distribution, and he wan…

题目链接 给一棵树, m个询问, 每个询问给出3个点, 求这三个点之间的最短距离. 其实就是两两之间的最短距离加起来除2. 倍增的lca模板 #include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <map> #include <s…

题目链接 : ZOJ Problem Set - 3195 题目大意: 求三点之间的最短距离 思路: 有了两点之间的最短距离求法,不难得出: 对于三个点我们两两之间求最短距离 得到 d1 d2 d3 那么最短距离就是 d = ( d1 + d2 + d3 ) / 2 要注意每个数组的范围大小,因为这个问题手抖敲错,TLE+RE一整页/(ㄒoㄒ)/~~ 用前向星来保存边和询问,空间卡的也很严 如下图所示:所求路线为紫色,等于蓝色+黄色+绿色之和的一半 代码: #include <bits/stdc…

题意:给定n个点,下面n-1行 u , v ,dis 表示一条无向边和边权值,这里给了一颗无向树 下面m表示m个询问,问 u v n 三点最短距离 典型的LCA转RMQ #include<stdio.h> #include<string.h> #include<math.h> #define N 100000 #define INF 1<<29 #define Logo 17 using namespace std; inline int Min(int a…

Design the city Time Limit: 1 Second      Memory Limit: 32768 KB Cerror is the mayor of city HangZhou. As you may know, the traffic system of this city is so terrible, that there are traffic jams everywhere. Now, Cerror finds out that the main reason…

这个题目大意是: 有N个城市,编号为0~N-1,给定N-1条无向带权边,Q个询问,每个询问求三个城市连起来的最小权值. 多组数据 每组数据  1 < N < 50000  1 < Q < 70000: 一道多源最短路的题目,注意题目数据:N.Q都很大 不能考虑Floyd.SPFA.Dijkstra.Bellman-Ford等最短路算法 再看N-1条边,明显构成一棵树,最短路有且只有一条 很明显需要LCA.... 不懂LCA的点!我!点!我! 我们所熟知的LCA是求两个点的最短路,而…

题目要对每次询问将一个树形图的三个点连接,输出最短距离. 利用tarjan离线算法,算出每次询问的任意两个点的最短公共祖先,并在dfs过程中求出离根的距离.把每次询问的三个点两两求出最短距离,这样最终结果就是3个值一半. 其实开始我用的一种很挫的方法才AC的,具体思路就不说了,感觉很麻烦又不好写的样子.怎么没想到上面的简便方法呢. 初始代码: #include <iostream> #include <sstream> #include <cstdio> #includ…

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…

Design the city Time Limit: 1 Second      Memory Limit: 32768 KB Cerror is the mayor of city HangZhou. As you may know, the traffic system of this city is so terrible, that there are traffic jams everywhere. Now, Cerror finds out that the main reason…

How far away ? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9359    Accepted Submission(s): 3285 Problem Description There are n houses in the village and some bidirectional roads connecting…

题目链接:传送门 题意: 给定一棵树,求两个点之间的距离. 分析: LCA 的模板题目 ans = dis[u]+dis[v] - 2*dis[lca(u,v)]; 在线算法:详细解说 传送门 代码例如以下: #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn = 40010; str…

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]] (…

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…

题目链接:http://poj.org/problem?id=1986 Description Farmer John's cows refused to run in his marathon since he chose a path much too long for their leisurely lifestyle. He therefore wants to find a path of a more reasonable length. The input to this prob…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2586 在线版本: 在线方法的思路很简单,就是倍增.一遍dfs得到每个节点的父亲,以及每个点的深度.然后用dp得出每个节点向上跳2^k步到达的节点. 那么对于一个查询u,v,不妨设depth[u]>=depth[v],先让u向上跳depth[u]-depth[v]步,跳的方法就是直接用数字的二进制表示跳. 然后现在u和v都在同一深度上了,再二分找向上共同的祖先,就可以二分出lca了.复杂度nlogn预…

解题关键:求树上三点间的最短距离. 解题关键:$ans = (dis(a,b) + dis(a,c) + dis(b,c))/2$ //#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<cmath> #i…

How far away ? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8712    Accepted Submission(s): 3047 Problem Description There are n houses in the village and some bidirectional roads connecting…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=2586 题意:给你N个点,M次询问.1~N-1行输入点与点之间的权值,之后M行输入两个点(a,b)之间的最近距离: 思路:设d[maxn]是当前点到根结点的距离,则答案就是 d[a]-d[lca(a,b)]+d[b]-d[lca(a,b)]: 接下来介绍LCA(Least Common Ancestors) 即最近公共祖先,是指在有根树中,找出某两个结点u和v最近的公共祖先. 常见解法一般有三种 这里讲…

查找最近公共祖先...我也不知道这东西有什么用,在线写法,非常之慢.... 存代码 #include<cstdio> #include<iostream> #include<cmath> #include<cstring> #include<algorithm> using namespace std; int n,m; struct nod{ int y; int next; }e[]; ]={}; ]={}; ]={}; ]={}; ]={}…

任意门:http://poj.org/problem?id=1330 Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 34942   Accepted: 17695 Description A rooted tree is a well-known data structure in computer science and engineering. An example…

以下转自:https://www.cnblogs.com/JVxie/p/4854719.html 首先是最近公共祖先的概念(什么是最近公共祖先?): 在一棵没有环的树上,每个节点肯定有其父亲节点和祖先节点,而最近公共祖先,就是两个节点在这棵树上深度最大的公共的祖先节点. 换句话说,就是两个点在这棵树上距离最近的公共祖先节点. 所以LCA主要是用来处理当两个点仅有唯一一条确定的最短路径时的路径. 有人可能会问:那他本身或者其父亲节点是否可以作为祖先节点呢? 答案是肯定的,很简单,按照人的亲戚观念…

<题目链接> 题目大意:给你一棵树,然后进行q次询问,然后要你统计这q次询问中指定的两个节点最近公共祖先出现的次数. 解题分析:LCA模板题,下面用的是离线Tarjan来解决.并且为了代码的简洁,本代码用的是vector存图. #include<iostream> #include<cstdio> #include<cstring> #include<vector> using namespace std; ; vector<int>…

<题目链接> 题目大意: 给出一棵树,问任意两个点的最近公共祖先的编号. 解题分析:LCA模板题,下面用的是树上倍增求解. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; struct Edge…

题目链接:https://www.luogu.org/problemnew/show/P3379 题意:LCA模板题. 思路:今天开始学树剖,先拿lca练练.树剖解lca,两次dfs复杂度均为O(n),每次查询为logn,因此总复杂度为:O(2*n+m*logn). 代码: #include<cstdio> #include<cstring> using namespace std; ; struct node{ int v,next; }edge[*maxn]; int n,m,…

How far away ? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 20971    Accepted Submission(s): 8245 Problem Description There are n houses in the village and some bidirectional roads connecting…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2586 题目: How far away ? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 21500    Accepted Submission(s): 8471 Problem Description There are n house…

Design the city Time Limit: 1 Second      Memory Limit: 32768 KB Cerror is the mayor of city HangZhou. As you may know, the traffic system of this city is so terrible, that there are traffic jams everywhere. Now, Cerror finds out that the main reason…

本文参考: https://www.cnblogs.com/GerynOhenz/p/8727415.html kuangbin的ACM模板(新) 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4347 Problem Description The course of Software Design and Development Practice is objectionable. ZLC is facing a serious problem…

因为徐州现场赛的G是树上差分+组合数学,但是比赛的时候没有写出来(自闭),背锅. 会差分数组但是不会树上差分,然后就学了一下. 看了一些东西之后,对树上差分写一点个人的理解: 首先要知道在树上,两点之间只有一条路径.树上差分就是在树上用差分数组,因为是在树上的操作,所以要用到lca,因为对于两点a,b,从a到b这一条链就是a-->lca(a,b)-->b,这是一条链. 其次,树上差分的两种操作:一种是对点权的,另一种是对边权的. 对于点权: 在树上将路径的起点a+1和终点b+1,lca(a,b…

C - Design the city Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Description Cerror is the mayor of city HangZhou. As you may know, the traffic system of this city is so terrible, that there are traffic jams everywhere.…