GTY's gay friends Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) [Problem Description] GTY has n gay friends. To manage them conveniently, every morning he ordered all his gay friends to stand in a line. Every gay…
3673: 可持久化并查集 by zky Time Limit: 5 Sec Memory Limit: 128 MBSubmit: 1878 Solved: 846[Submit][Status][Discuss] Description n个集合 m个操作操作:1 a b 合并a,b所在集合2 k 回到第k次操作之后的状态(查询算作操作)3 a b 询问a,b是否属于同一集合,是则输出1否则输出0 0<n,m<=2*10^4 Input Output Sample Input 5 6…
题意: 把手臂都各自看成一个向量,则机械手的位置正好是手臂向量之和.旋转某个关节,其实就是把关节到机械手之间的手臂向量统统旋转. 由于手臂很多,要每个向量做相同的旋转操作很费时间.这时就可以想到用线段树的优势正是可以快速地成段更新.和一般的成段更新题目没什么差别,只是通常成段替换.或者成段增加.这时候要做的是,对向量成段得做旋转变换.只要利用旋转变化矩阵即可. 要注意,从第 s 节手臂开始到机械手要旋转的角度,和第 s - 1 节手臂的角度有关.因此我们还要记录每个手臂的角度,并且去Query得…
题目链接 题意: 给定一棵树,每条边有黑白两种颜色,初始都是白色,现在有三种操作: 1 u v:u到v路径(最短)上的边都取成相反的颜色 2 u v:u到v路径上相邻的边都取成相反的颜色(相邻即仅有一个节点在路径上) 3 u v:查询u到v路径上有多少个黑色边 思路: 对树进行树链剖分,分成重链和轻链,用两棵线段树W,L来维护.W维护树上在重链上的u和v之间的边的翻转情况(操作在线段树上的[pos[v],pos[u]]区间):L维护树上在重链上的u和v之间的相邻边的翻转情况.那么某一个点u与它父…
FZU 2105 Digits Count Time Limit:10000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Practice Description Given N integers A={A[0],A[1],...,A[N-1]}. Here we have some operations: Operation 1: AND opn L R Here opn, L and R are intege…
ZYB has a premutation P,but he only remeber the reverse log of each prefix of the premutation,now he ask you to restore the premutation. Pair (i, j)(i < j) is considered as a reverse log if Ai > Aj is matched.Input In the first line there is the num…
题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=5475 Problem Description One day, a useless calculator was being built by Kuros. Let's assume that number X is showed on the screen of calculator. At first, X = 1. This calculator only supports two types…
C. Sereja and Brackets time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Sereja has a bracket sequence s1, s2, ..., sn, or, in other words, a string s of length n, consisting of characters "(…