有一棵n个节点的树,一共q 次询问 每次询问给定3个点,求两条起点终点在这三个点上且不完全重合的路径的最多公共节点数 简单LCA求距离,令a为汇合点,那么答案就是(dis(a,b) + dis(a,c) - dis(b,c)) / 2 + 1,dis用lca求出,枚举a就好. 当然也可以一一讨论abc的位置关系,不过容易出错. #include<queue> #include<cstdio> #include<iostream> #include<algorith…

832D - Misha, Grisha and Underground 思路:lca,求两个最短路的公共长度.公共长度公式为(d(a,b)+d(b,c)-d(a,c))/2. 代码: #include<bits/stdc++.h> using namespace std; #define ll long long #define ls rt<<1,l,m #define rs rt<<1|1,m+1,r const int INF=0x3f3f3f3f; ; ; ve…

Misha, Grisha and Underground time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Misha and Grisha are funny boys, so they like to use new underground. The underground has n stations connected…

Misha, Grisha and Underground 题意:Misha 和 Grisha 是2个很喜欢恶作剧的孩子, 每天早上 Misha 会从地铁站 s 通过最短的路到达地铁站 f, 并且在每个地铁站上都写上一句话, 然后Grisha 再从地铁站 t 通过最短的路到达地铁站 f, 并且记录下路途上被Misha写上字的地铁站数目,并且当天晚上会人会将地铁站清理干净,不会干扰第二天的计数, 现在给你3个地铁站 a, b, c, 现在求Misha记录的数目最大能是多少. 代码:求出Lca(a,…

D. Misha, Grisha and Underground time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Misha and Grisha are funny boys, so they like to use new underground. The underground has n stations connec…

D. Misha, Grisha and Underground 这个题目算一个树链剖分的裸题,但是这个时间复杂度注意优化. 这个题目可以选择树剖+线段树,时间复杂度有点高,比较这个本身就有n*logn*logn 但是就是lca+一点点思维就完全不卡时间. #include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <algorithm&g…

Misha and Grisha are funny boys, so they like to use new underground. The underground has n stations connected with n - 1 routes so that each route connects two stations, and it is possible to reach every station from any other. The boys decided to h…

Misha and Grisha are funny boys, so they like to use new underground. The underground has n stations connected with n - 1 routes so that each route connects two stations, and it is possible to reach every station from any other. The boys decided to h…

我奇特的脑回路的做法就是 树链剖分 + 树状数组 树状数组是那种 区间修改,区间求和,还有回溯的 当我看到别人写的是lca,直接讨论时,感觉自己的智商收到了碾压... #include<cmath> #include<map> #include<iostream> #include<cstring> #include<cstdio> #include<set> #include<vector> #include<qu…

一棵树,q次询问,每次给你三个点a b c,让你把它们选做s f t,问你把s到f +1后,询问f到t的和,然后可能的最大值是多少. 最无脑的想法是链剖线段树……但是会TLE. LCT一样无脑,但是少一个log,可以过. 正解是分类讨论, 如果t不在lca(s,f)的子树内,答案是dis(lca(s,f),f). 如果t在lca(s,f)的子树内,并且dep(lca(s,t))>dep(lca(f,t)),答案是dis(lca(s,t),f): 否则答案是dis(lca(f,t),f). #in…

[Link]:http://codeforces.com/contest/832/problem/D [Description] 给你一棵树; 然后给你3个点 让你把这3个点和点s,t,f对应; 然后s先从s走到f; 之后t再从t走到f; 求这两条路径的公共路径的长度; [Solution] 答案为 dis(s,f)+dis(t,f)−dis(s,t)2 树上最短路径做一下就好; LCA! [NumberOf WA] 0 [Reviw] 想得太慢了 [Code] #include <cstdio…

题目链接 模板copy from http://codeforces.com/contest/832/submission/28835143 题意,给出一棵有n个结点的树,再给出其中的三个结点 s,t,f ,求路径 (s,t) (s,f) (t,f) 三者的最大公共结点数 对于(t,f) (s,f)的公共结点数,有 懒得细想了,暴力对s,t,f生成排列(反正一共仨数,最多也就3!=6个排列),求各种情况下的最大ans #include<bits/stdc++.h> using namespac…

题意:在一棵生成树上,给出了三个点,求三个点之间最大的相交点数,CF难度1900. 题解:求出三个lca,并取深度最大的那个,就是我们要的三岔路口K,然后分别求出K到a,b,c三点的路径长度,取最大值+1就是答案.为什么是取深度最大的呢?因为只有当深度是最大的时候设该点为K,这个K为三岔路口,每一个a,b,c到K的距离都不会用重复,否则如果取的不是最深的,可能造成重复的情况,这个需要避免.然后找到这个K之后,ans就是a,b,c三点分别到K的距离+1即可(+1是因为本身也算). 一开始没做出来的…

A. Sasha and Sticks 题目链接:http://codeforces.com/contest/832/problem/A 题目意思:n个棍,双方每次取k个,取得多次数的人获胜,Sasha先取,问Sasha是否可以取胜. 代码: //Author: xiaowuga #include <iostream> #include <algorithm> #include <set> #include <vector> #include <que…

A. Lorenzo Von Matterhorn B.Minimum spanning tree for each edge C.Misha, Grisha and Underground D.Fools and Roads E.City Driving 题意:给你一颗基环树(有n条边,n个点的连通图),有Q个询问,求u,v之间的距离最小是多少 思路:对于一棵基环树而言,只要去掉其环上任意一条边(a , b),其边权为w ,剩下的边就可以构成一棵树 对于 u,v 之间的最小距离 , 有可能由去…

By the Underground or by Foot? Time limit: 1.0 secondMemory limit: 64 MB Imagine yourself in a big city. You want to get from point A to point B. To do that you may move by foot or use the underground. Moving by the underground is faster but you may…

C. Misha and Forest   Let's define a forest as a non-directed acyclic graph (also without loops and parallel edges). One day Misha played with the forest consisting of n vertices. For each vertex v from 0 to n - 1 he wrote down two integers, degreev …

题目链接: 题目 Underground Cables Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu 问题描述 A city wants to get rid of their unsightly power poles by moving their power cables underground. They have a list of points that all need to be con…

题目链接: 题目 A. Misha and Forest time limit per test 1 second memory limit per test 256 megabytes 问题描述 Let's define a forest as a non-directed acyclic graph (also without loops and parallel edges). One day Misha played with the forest consisting of n ver…

About Grisha N. 题目链接: http://acm.hust.edu.cn/vjudge/contest/126546#problem/A Description Grisha N. told his two teammates that he was going to solve all given problems at the quarter-finals, even if all his teammates wouldn't show up at the competiti…

Description Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used a…

Misha and Palindrome Degree 题目链接:http://codeforces.com/problemset/problem/501/E 贪心 如果区间[L,R]满足条件,那么区间[L',R'](L'<=L,R<=R')必然满足条件,所以只需要找到满足条件的最小区间即可.首先去除两边相同的区间,剩下的区间为[l,r],因为区间[l,r]的两端不相同,所以要找的最小区间必然包含区间[l,r]的最左端或者最右端.观察到所选区间内的同种元素个数必需大于等于整个区间内同种元素的个…

Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point…

E. Misha and LCP on Tree Problem's Link Mean: 给出一棵树,每个结点上有一个字母.每个询问给出两个路径,问这两个路径的串的最长公共前缀. analyse: 做法:树链剖分+后缀数组. 记录每条链的串,正反都需要标记,组成一个长串. 然后记录每条链对应的串在大串中的位置,对大串求后缀数组,最后询问就是在一些链上的查询. Time complexity: O(n*logn) view code ));    );    ; ; ; ; ) ;      …

地址:http://codeforces.com/contest/782/problem/E 题目: E. Underground Lab time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The evil Bumbershoot corporation produces clones for gruesome experimen…

本文版权归ljh2000和博客园共有,欢迎转载,但须保留此声明,并给出原文链接,谢谢合作. 本文作者:ljh2000 作者博客:http://www.cnblogs.com/ljh2000-jump/转载请注明出处,侵权必究,保留最终解释权! 题目链接:codeforces781C Underground Lab 正解:生成树+欧拉序列 解题报告: 很容易发现,原图和图的一棵生成树其实是等价的,都是需要遍历全图,是一个类似于$dfs$的东西. 欧拉序列正可以满足题目所给的性质,所以我只要搞出一棵…

Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point…

2012. About Grisha N. Time limit: 1.0 secondMemory limit: 64 MB Grisha N. told his two teammates that he was going to solve all given problems at the subregional contest, even if the teammates wouldn’t show up at the competition. The teammates didn’t…

Misha and Changing Handles CodeForces原题是英文,这里就直接上中文好了,翻译不是太给力,但是不影响做题 ^▽^ Description  神秘的三角洲里还有一个传说中的谜题等你来解开!三角洲里的小学生小明是个小天才,知天文晓地理还能夜观星象算命,好多疯狂的小朋友都想去请他给自己换个好听的名字.但是天才小明,他总是在思考31维宇宙空间的奥秘,神游天外,所以在给小朋友们敲他们想要的名字的时候,偶尔会取出一些不那么完美的名字.有的小朋友们换了名字以后不太满意,于是它…

题目传送门 /* 题意:给出无向无环图,每一个点的度数和相邻点的异或和(a^b^c^....) 图论/位运算:其实这题很简单.类似拓扑排序,先把度数为1的先入对,每一次少一个度数 关键在于更新异或和,精髓:a ^ b = c -> a ^ c = b, b ^ c = a; */ #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <…