Little Petya likes points a lot. Recently his mom has presented him n points lying on the line OX. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed d. Not…

原文地址 Python+Arcpy操作Points(.shp)转换至Polyline(.shp),仔细研读Points To Line (Data Management)说明,参数说明如下: Input_Features: The point features to be converted into lines. Output_Feature_Class:The line feature class which will be created from the input points. 以下…

Psychos in a Line CodeForces - 319B There are n psychos standing in a line. Each psycho is assigned a unique integer from 1 to n. At each step every psycho who has an id greater than the psycho to his right (if exists) kills his right neighbor in the…

A. Points on Line time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Little Petya likes points a lot. Recently his mom has presented him n points lying on the line OX. Now Petya is wondering…

Description Little Petya likes points a lot. Recently his mom has presented him n points lying on the line OX. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't e…

Two Sum: 解法一:排序后使用双索引对撞:O(nlogn)+O(n) = O(nlogn) , 但是返回的是排序前的指针. 解法二:查找表.将所有元素放入查找表, 之后对于每一个元素a,查找 target-a 是否存在.使用map实现,键是元素的值,键值是元素对应的索引. 不能把vector中所有的值放到查找表中,因为若有重复的值,前一个会被后一个覆盖.所以改善为把当前元素v前面的元素放到查找表中. 时间复杂度:O(n) 空间复杂度:O(n) 注意:这道题只有唯一解. class Solu…

Little Petya likes points a lot. Recently his mom has presented him n points lying on the line OX. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed d. Not…

题目链接: https://vjudge.net/problem/35188/origin 题目大意: 要求你找到一个 i < j < k时有 a[k]-a[i] <= d的组的个数. 这道题的思路就是每输入一个a[i]就是找到那个第一个大于或等于a[i]-d的值!,设a[i]-d = mn, 则用lower_bound()函数可以得到这个第一个大于或者等于mn的值的下标. 关于lower_bound()函数的详细用法可以看这个博客: https://blog.csdn.net/qq_4…

题解: 方法非常巧妙的一道题 首先考虑要求全部为0怎么做 发现是个欧拉回路的问题(很巧妙) 直接dfs一遍就可以了 而这道题 要求是-1,1,0 我们可以先离散化 完了之后判断每个点被奇数还是偶数条边覆盖 如果是奇数,那么就多连一条边 另外有个细节是为了要用欧拉回路区间左开右闭 代码: #include <bits/stdc++.h> using namespace std; const int N=4e5; ; int n,cnt[N],rd[N],b[N],head[N],l1; stru…

题意:有n个点,找到一个顺序走遍这n个点,并且曼哈顿距离不超过25e8. 由于给的点坐标都在0-1e6之间.将x轴分成1000*1000,即1000长度为1块.将落在同一块的按y排序,编号为奇的块和偶的块一个升序,一个降序.有3个量值得关注.一是同一块中x的变化量,其实是不超过1000*n1,n1是第1块中点的数量.那么1000*n1+1000*n2......=1000*n<1e9.后两个量是同一块中y的高度差,另一个是本块最后一个和另一块第一个的高度差.这种做法下相邻两块这两个高度差的和是小…