#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #define maxn 2000 using namespace std; const double pi=acos(-1.0); double sqr(double x) { return x*x; } struct node { double x,y; }p[maxn]; double dis(node…

Rope in the Labyrinth Time Limit:500MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice URAL 1145 Description A labyrinth with rectangular form and size m × n is divided into square cells with sides' length 1 by lines…

1145. Rope in the Labyrinth Time limit: 0.5 secondMemory limit: 64 MB A labyrinth with rectangular form and size m × n is divided into square cells with sides' length 1 by lines that are parallel with the labyrinth's sides. Each cell of the grid is e…

1246. Tethered Dog Time limit: 1.0 secondMemory limit: 64 MB A dog is tethered to a pole with a rope. The pole is located inside a fenced polygon (not necessarily convex) with nonzero area. The fence has no self-crosses. The Olympian runs along the f…

POJ 上的一套水题,哈哈~~~,最后一题很恶心,不想写了~~~ Rope Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7410   Accepted: 2603 Description Plotters have barberically hammered N nails into an innocent plane shape, so that one can see now only heads. Moreove…

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 Sample Output Solution 用rope实现可持久化数组,用rope的历史记录功能实现可持久化并查集,通过时间168ms #include<cstdio> #include<ext/rop…

参考资料 1)官方说明 支持 sorry,cena不支持rope 声明 1)头文件 #include<ext/rope> 2)调用命名空间 using namespace __gnu_cxx; 底层原理 查了资料,大概可以称作可持久化平衡树,因为rope适用于大量.冗长的串操作,而不适合单个字符操作官方说明如下: Though ropes can be treated as Containers of characters, and are almost Sequences, this is…

Description 这些日子,可可不和卡卡一起玩了,原来可可正废寝忘食的想做一个简单而高效的文本编辑器.你能帮助他吗?为了明确任务目标,可可对“文本编辑器”做了一个抽象的定义:   文本:由0个或多个字符构成的序列.这些字符的ASCII码在闭区间[32, 126]内,也就是说,这些字符均为可见字符或空格.光标:在一段文本中用于指示位置的标记,可以位于文本的第一个字符之前,文本的最后一个字符之后或文本的某两个相邻字符之间.文本编辑器:为一个可以对一段文本和该文本中的一个光标进行如下七条操作的程…

题目链接 题意:求给定的字符串的最长回文子串 分析:做法是构造一个新的字符串是原字符串+反转后的原字符串(这样方便求两边回文的后缀的最长前缀),即newS = S + '$' + revS,枚举回文串中心位置,RMQ询问LCP = min (height[rank[l]+1] to height[rank[r]]),注意的是RMQ传入参数最好是后缀的位置,因为它们在树上的顺序未知,且左边还要+1. #include <cstdio> #include <algorithm> #in…

2071. Juice Cocktails Time limit: 1.0 secondMemory limit: 64 MB Once n Denchiks come to the bar and each orders a juice cocktail. It could be from 1 to 3 different juices in each cocktail. There are three juices in the bar: apple, banana and pineappl…