题目链接: E. Connecting Universities time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Treeland is a country in which there are n towns connected by n - 1 two-way road such that it's possible to…

题目链接:http://codeforces.com/contest/701/problem/E 题意:有n个城市构成一棵树,一个城市最多有一个学校,这n个城市一共2*k个学校,要对这2*k个学校进行连边,使得所有连出来的边的和最大,每条边边权为1. 题解:这题有一个巧妙的解法,可以记录一下每一条边的贡献(所谓贡献就是有多少对点连线会经过这条边) 也就是记录一下这条边左端点以左的所有需要连的点x和右端点的数目就是2*k-x.然后边的贡献就是两者取 最小.这样dfs一遍所有的边就行了. #incl…

Descrition 给你一颗\(n\le 2*10^5\)个点的树, 有\(2*k(2k\le n)\)座大学座落在点上 (任二大学不在同一个点) 求一种两两匹配的方案, 使得距离和最大 即\[maximize~\{~\sum_{each~pair~(x,y)} dis(x,y)~\}~\] Solution 1 (1) 化简一下我们相当于要最小化 两两lca的深度和 我们先把这2k所大学按dfn序从小到大排好, 把前k个称为A部分, 后k个称为B部分 (2) 所有匹配均为\(A-B\)匹配…

初始的时候有一个只有n个点的图(n <= 1e5), 现在进行m( m <= 1e5 )次操作 每次操作要么添加一条无向边, 要么询问之前结点u和v最早在哪一次操作的时候连通了 /* * Author: Gatevin * Created Time: 2015/11/21 14:02:38 * File Name: Sakura_Chiyo.cpp */ #include<iostream> #include<sstream> #include<fstream&g…

链接 Codeforces 701E Connecting Universities 题意 n个点的树,给你2*K个点,分成K对,使得两两之间的距离和最大 思路 贪心,思路挺巧妙的.首先dfs一遍记录每个点的子树中(包括自己)有多少点是这K个中间的,第二遍dfs时对于每一条边,取两端包含较少的值,这样就保证树中间的点不会被取到,留下的就是相隔更远的点了.方法确实想不到啊. 代码 #include <iostream> #include <cstdio> #include <v…

E. Connecting Universities time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Treeland is a country in which there are n towns connected by n - 1 two-way road such that it's possible to get f…

E. Connecting Universities time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Treeland is a country in which there are n towns connected by n - 1 two-way road such that it's possible to get f…

Connecting Universities Treeland is a country in which there are n towns connected by n - 1 two-way road such that it's possible to get from any town to any other town. In Treeland there are 2k universities which are located in different towns. Recen…

pid=5647">[HDU 5647]DZY Loves Connecting(树DP) DZY Loves Connecting Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 332    Accepted Submission(s): 112 Problem Description DZY has an unroote…

Buses and People CodeForces 160E 三维偏序+线段树 题意 给定 N 个三元组 (a,b,c),现有 M 个询问,每个询问给定一个三元组 (a',b',c'),求满足 a<a', b'<b, c'<c 的最小 c 对应的元组编号. 解题思路 三维偏序问题,是我第一次做,查的题解. 一位大佬是这么说的,原博客首先,离线处理所有询问,将这 N+M 个元组按照 a 从小到达进行排序,若有相同的 a,则给定元组应该排在询问元组之前.排序后即可保证对于任意一个询问元组…