LCA问题的tarjan解法模板 LCA问题 详细 1.二叉搜索树上找两个节点LCA public int query(Node t, Node u, Node v) { int left = u.value; int right = v.value; //二叉查找树内,如果左结点大于右结点,不对,交换 if (left > right) { int temp = left; left = right; right = temp; } while (true) { //如果t小于u.v,往t的右…

POJ 1470 Closest Common Ancestors(最近公共祖先 LCA) Description Write a program that takes as input a rooted tree and a list of pairs of vertices. For each pair (u,v) the program determines the closest common ancestor of u and v in the tree. The closest co…

任意门:http://poj.org/problem?id=1470 Closest Common Ancestors Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 22519   Accepted: 7137 Description Write a program that takes as input a rooted tree and a list of pairs of vertices. For each pa…

Closest Common Ancestors Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 13372   Accepted: 4340 Description Write a program that takes as input a rooted tree and a list of pairs of vertices. For each pair (u,v) the program determines the…

传送门 Closest Common Ancestors Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 17306   Accepted: 5549 Description Write a program that takes as input a rooted tree and a list of pairs of vertices. For each pair (u,v) the program determines…

Closest Common Ancestors Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 13370   Accepted: 4338 Description Write a program that takes as input a rooted tree and a list of pairs of vertices. For each pair (u,v) the program determines the…

Closest Common Ancestors Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 20804   Accepted: 6608 Description Write a program that takes as input a rooted tree and a list of pairs of vertices. For each pair (u,v) the program determines the…

版权声明:本文为博主原创文章,未经博主同意不得转载. https://blog.csdn.net/u013912596/article/details/35311489 题目链接:http://poj.org/problem?id=1470 题目大意:给出一棵树.再给出若干组数(a,b),输出节点a和节点b的近期公共祖先(LCA) 就是非常裸的LCA.可是我用的是<挑战程序设计竞赛>上的"基于二分搜索的算法求LCA",我看网上用的都是tarjan算法.可是我的代码不知道为什…

Tarjan算法的详细介绍,请戳: http://www.cnblogs.com/chenxiwenruo/p/3529533.html #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <string> #include <vector> /* AC 一开始读取数据的方式并不好,运行900多ms. 后来参照…

题目链接:http://poj.org/problem?id=1470 Write a program that takes as input a rooted tree and a list of pairs of vertices. For each pair (u,v) the program determines the closest common ancestor of u and v in the tree. The closest common ancestor of two n…

1.输出测试用例中是最近公共祖先的节点,以及这个节点作为最近公共祖先的次数. 2.最近公共祖先,离线Tarjan算法 3. /* POJ 1470 给出一颗有向树,Q个查询 输出查询结果中每个点出现次数 */ /* 离线算法,LCATarjan 复杂度O(n+Q); */ #include<iostream> #include<stdio.h> #include<string.h> using namespace std; ; ;//查询数的最大值 //并查集部分 in…

<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>…

A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:  In the figure, each node is labeled with an integer from {1, 2,...,16}. Node 8 is the root of the tree. Node x is an ancestor of node y if n…

其实这是一个裸求LCA的题目,我使用的是离线的Tarjan算法,但是这个题的AC对于我来说却很坎坷……首先是RE,我立马想到数组开小了,然后扩大了数组,MLE了……接着把数组调整适当大小,又交了一发,嗯?居然WA了,一定是我姿势不对,我换个编译器交一下,结果还是WA……这就比较尴尬了,最后审题目,发现了,我以为第一个输入的点就是根节点,其实不是,我们还需要找到入度为0的点来处理,这里做一个标记就可以了.还有就是题目的输入稍微有点坑,尤其是第二次输入,题目中说忽略所有空白,这就需要比较好的处理方法…

题目链接: http://poj.org/problem?id=1470 题意: 给定若干有向边,构成有根数,给定若干查询,求每个查询的结点的LCA出现次数. 分析: 还是很裸的tarjan的LCA. 这题我真的要吐槽!!调试了好久啊!!不是MLE就是RE.... 查询数量没给,所以要用矩阵来存,这样还可以避免重复. 给边的时候不知道会不会有重复的点,所以讲道理应该用vector保存. getchar...我不知道为什么会RE... 其实ance数组可以不用的,只要每次处理子树的时候pa[v]…

题意比较费劲:输入看起来很麻烦.处理括号冒号的时候是用%1s就可以.还有就是注意它有根节点...Q次查询 在线st算法 /************************************************************************* > File Name: 3.cpp > Author: Howe_Young > Mail: 1013410795@qq.com > Created Time: 2015年10月08日 星期四 19时03分30秒…

本题也是找LCA的题目,只是要求多次查询.一般的暴力查询就必定超时了,故此必须使用更高级的方法,这里使用Tarjan算法. 本题处理Tarjan算法,似乎输入处理也挺麻烦的. 注意: 由于查询的数据会极大,故此使用一个数组记录全部查询数据就会超时的. 我就载在这里了.查了好久才想到这点. 由于我使用了一个vector容器记录了查询数据.故此每次都循环这组这么大的数据,就超时了. ----解决的方法:使用一个vector<int> quest来记录查询数组.这样每次都仅仅须要循环某节点的邻接查询…

用的离线算法Tarjan 该算法的详细解释请戳 http://www.cnblogs.com/Findxiaoxun/p/3428516.html 做这个题的时候,直接把1470的代码copy过来,改了改输入输出.这个的难度比那个低. #include<cstdio> #include<algorithm> #include<cstring> #include<vector> using namespace std; ; int father[MAXN],a…

该算法的详细解释请戳: http://www.cnblogs.com/Findxiaoxun/p/3428516.html #include<cstdio> #include<algorithm> #include<cstring> #include<vector> using namespace std; ; int father[MAXN],ancestor[MAXN]; bool visit[MAXN]; int ans[MAXN]; vector&l…

POJ 1330 Nearest Common Ancestors 题意:最近公共祖先的裸题 思路:LCA和ST我们已经很熟悉了,但是这里的f[i][j]却有相似却又不同的含义.f[i][j]表示i节点的第2j个父亲是多少   这个代码不是我的,转自 邝斌博客 /* *********************************************** Author :kuangbin Created Time :2013-9-5 9:45:17 File Name :F:\2013AC…

http://poj.org/problem?id=1470 Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 20830   Accepted: 6617 Description Write a program that takes as input a rooted tree and a list of pairs of vertices. For each pair (u,v) the program determin…

Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 18136   Accepted: 9608 Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: In the figure, each…

A - Nearest Common Ancestors Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%lld & %llu Submit Status Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: In the figu…

Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 37386   Accepted: 18694 Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: In the figure, each…

http://poj.org/problem? id=1330 给一个有根树,一个查询节点(u,v)的近期公共祖先 836K 16MS #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<vector> #include<string> #include<set> #include<map> con…

LCA思想:http://www.cnblogs.com/hujunzheng/p/3945885.html 在求解最近公共祖先为问题上,用到的是Tarjan的思想,从根结点开始形成一棵深搜树,非常好的处理技巧就是在回溯到结点u的时候,u的子树已经遍历,这时候才把u结点放入合并集合中,这样u结点和所有u的子树中的结点的最近公共祖先就是u了,u和还未遍历的所有u的兄弟结点及子树中的最近公共祖先就是u的父亲结点.以此类推..这样我们在对树深度遍历的时候就很自然的将树中的结点分成若干的集合,两个集合中…

传送门 Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 26612   Accepted: 13734 Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:    In the figur…

<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…

非常裸的模版题,只是Tarjan要好好多拿出来玩味几次 非常有点巧妙呢,tarjan,大概就是当前结点和它儿子结点的羁绊 WA了俩小时,,,原因是,这个题是多数据的(还没告诉你T,用scanf!=EOF来控制结束),更重要的是和这个和Codeforces不一样,Codeforces的多组数据好像会又一次開始程序似的,不用在程序里面写清零,但这个题是多数据用EOF来控制输入的,多数据在一个文件中都一次输进去了,所以要memset btw,加上一点memset代码,多了700B代码... #incl…

POJ 1330 Nearest Common Ancestors / UVALive 2525 Nearest Common Ancestors (最近公共祖先LCA) Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: In the figure, each node is labeled with an…