Search - plane points

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[Data structs] POJ1113 DL : 0: 问题：已经平面上的若干点，求包围这些点的最小包围长度算法：经典的凸包算法，注意求解时对最外圈做处理。-Question: has a number of points on the plane, and surrounded by the smallest of these points, surrounded by the length of algorithm: the classic convex hull algorithm, when attention to solving the most to do deal with the outer ring.
Date : 2025-12-29 Size : 1kb User : 苏钿煌
[Data structs] zjdd DL : 0: 用于计算二维平面上一个点集的最近点对，可以计算100000个点的规模-Used to calculate the two-dimensional plane, a set of points the last points, we can calculate the size of 100000 points
Date : 2025-12-29 Size : 2kb User : liam
[Data structs] dbPoint.c DL : 0: C语言实现，分治法求平面中任意个点的最近点对和最近距离。-C language implements, find out the shortest point pair and the shortest length between arbitrary points in a plane, with divide and conquer method.
Date : 2025-12-29 Size : 1kb User : liuwei
[Data structs] CLOSESTPAIR DL : 0: 最近点对问题，求解平面上n个点的集合中距离最近的两点的距离。-Nearest point of the problem, solving the plane the set of n points in the distance between the nearest two points.
Date : 2025-12-29 Size : 1kb User : hyqin
[Data structs] PointPair DL : 0: 根据分治算法实现求平面上最近点对的复杂度为(nlgn)的算法有图形界面，能通过鼠标输入点-Under sub-rule algorithm seeking the nearest point on the plane of the complexity of (nlgn) algorithm for graphical interface, through mouse input points
Date : 2025-12-29 Size : 740kb User : liupz
[Data structs] JudgePointInRect DL : 0: 根据矩形左边中点、方向及长宽求矩形各点坐标；判断平面上一点是否在该平面上某一不规则矩形内-According to the midpoint of the rectangle on the left, direction and length and width of a rectangular coordinate points determine whether the plane a little irregular in the rectangular plane within a
Date : 2025-12-29 Size : 49kb User : haochaoqing
[Data structs] closepair DL : 0: 二位平面中求最近点对的问题，采用分治的方法，按照书本的算法实现-Two nearest points on the plane ask the question, the use of divide and conquer approach, in accordance with the books of the algorithm
Date : 2025-12-29 Size : 449kb User : 龙飞
[Data structs] 1.1 DL : 0: C++设计一个表示平面直角坐标系的点的位置Location类，提供函数得到该点的坐标、计算两个点之间的距离，并且能够自动统计坐标系中点的个数。测试要求：在主程序中创建两个对象A和B，按如下格式输出两个点的坐标和两个点的距离。 A(x1, y1), B(x2, y2), Distance=d -C++ to design a plane rectangular coordinate system, said the positions of points Location class provides functions to get the coordinates of points, calculate the distance between two points, and automatically the number of statistical mid-point coordinates. Test requirements: in the main program creates two objects A and B, two points in the following format output coordinates and the distance between two points. A (x1, y1), B (x2, y2), Distance = d
Date : 2025-12-29 Size : 11kb User : xj
[Data structs] TSP DL : 0: 　欧氏旅行售货员问题是对给定的平面上n 个点确定一条连接这n 个点的长度最短的哈密顿回路。由于欧氏距离满足三角不等式，所以欧氏旅行售货员问题是一个特殊的具有三角不等式性质的旅行售货员问题。它仍是一个NP 完全问题。最短双调TSP 回路是欧氏旅行售货员问题的特殊情况。平面上n 个点的双调TSP 回路是从最左点开始，严格地由左至右直到最右点，然后严格地由右至左直至最左点，且连接每一个点恰好一次的一条闭合回路。-Euclidean traveling salesman problem is a given n points in the plane of a connection to determine the length of these n points the shortest Hamiltonian circuit. Euclidean distance to satisfy the triangle inequality, so the Euclidean traveling salesman problem is a special character with the triangle inequality the traveling salesman problem. It is still a NP complete problem. Double adjustable loop is the shortest Euclidean TSP traveling salesman problem a special case. N points in the plane two-tone TSP from the left point of the loop is started, the most strictly from left to right until the right point, and strictly from right to left until the left point, and connecting each point exactly once in a closed circuit.
Date : 2025-12-29 Size : 2kb User : viewhao
[Data structs] Kruskal DL : 0: 编程实现Kruskal算法，生成最小代价生成树，其中利用最小堆算法实现。（随机生成n个点，且随机生成k条边，形成连通图）根据输入的顶点数的不同，分析时间复杂度。-Implement Kruskal’s algorithm based on min-heap and disjoint set data structure for constructing a minimum cost spanning tree. Generate weighted undirected complete graph for given size n according to the following approach: 1) Randomly generate n points (x,y) on a plane 2) Randomly generate k pairs of points as edges of the graph. 3) The weight of an edge of the graph is defined as the the Euclidean distance between two endpoints of the edge.
Date : 2025-12-29 Size : 2kb User : 许许
[Data structs] Prim DL : 0: 编程实现Prim算法，基于最小堆数据结构，生成最小代价生成树。（其中随机生成点和边，形成连通图）根据输入的顶点数的不同，分析时间复杂度。-Implement Prim’s algorithms based on min-heap and disjoint set data structure for constructing a minimum cost spanning tree. Generate weighted undirected complete graph for given size n according to the following approach: 1) Randomly generate n points ( ) on a plane 2) Randomly generate k pairs of points as edges of the graph. 3) The weight of an edge of the graph is defined as the the Euclidean distance between two endpoints of the edge.
Date : 2025-12-29 Size : 2kb User : 许许
[Data structs] 11 DL : 0: 给定平面上n个点，找其中的一对点，使得在n个点组成的所有点对中，该点对间的距离最小。-Given n points on the plane, find a pair of points which makes the composition of n points All points, the distance between the point of minimum.
Date : 2025-12-29 Size : 38kb User : 王海颖081
[Data structs] asd DL : 0: 欧氏旅行售货员问题是对给定的平面上n 个点确定一条连接这n 个点的长度最短的哈密顿回路。-Euclidean traveling salesman problem is a given n points in the plane of a connection to determine the length of these n points the shortest Hamiltonian circuit.
Date : 2025-12-29 Size : 1kb User : 李书沙
[Data structs] cordic-c-vc-6.0 DL : 0: cordic算法的c语言实现，输入x，y两点坐标，自动计算得出改点与平面坐标轴（x轴正向）夹角，适用于整个坐标平面。于环境vc 6.0下编译通过。-CORDIC algorithm realize the C language, input x, y coordinates of two points, are automatically calculated to change point and plane coordinate axis ( x axis positive ) angle, apply to the whole coordinate plane. In the environment of VC 6 under the compiler through.
Date : 2025-12-29 Size : 2kb User : baijingdong
[Data structs] cordic-fixed-point-c-vc6.0- DL : 0: cordic算法的c语定点数仿真，输入x，y两点坐标，自动计算得出改点与平面坐标轴（x轴正向）夹角（有一定误差），适用于整个坐标平面。于环境vc 6.0下编译通过。-CORDIC algorithm C language fixed point simulation, enter the X, y coordinates of two points, are automatically calculated to change point and plane coordinate axis ( x axis angle ( positive ) has a certain error ), applicable to the coordinate plane. In the environment of VC 6 under the compiler through.
Date : 2025-12-29 Size : 2kb User : baijingdong
[Data structs] CLOSESTPAIR DL : 0: 给定平面上的N个点，找出两个点使其间距最小，并求出其距离。-Given N points on the plane to identify the two points so that the minimum spacing to determine its distance.
Date : 2025-12-29 Size : 1kb User : 马骏
[Data structs] MaxPointsonaLine DL : 0: Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 分析：首先要注意的是，输入数组中可能有重复的点。由于两点确定一条直线，一个很直观的解法是计算每两个点形成的直线，然后把相同的直线合并，最后包含点最多的直线上点的个数就是本题的解。我们知道表示一条直线可以用斜率和y截距两个浮点数（垂直于x轴的直线斜率为无穷大，截距用x截距），同时还需要保存每条直线上的点（避免重复）。听起来就很麻烦，但是基于这个思想有一种简单的实现方式：以某点O为中心，计算它和其他点的斜率，如果有两个点A、B和O点形成的斜率相等，那么ABO三点是共线的，如果有多个点和O的斜率相等，那么这多个点和O也是共线的，因此我们可以求出包含O的所有直线中点数最多的直线，会得到一个最大共线点数k(O)，如果和O点重复的点有n个（除了O本身），那么K(O) = K(O) + n。这一步的计算中，为了提高效率，我们可以用哈希表来保存某个斜率对应的点的数目。对数组中的每一个点i，按照第一步计算得到每个点最大点数K(i) 从k(i)中选取最大的就是本题的解注意：为了避免重复计算，以数组中第i个点为中心时，只计算数组中它右边的所有点-Given n points on a 2D plane, find the maximum number of points that lie on the same straight line analysis: first thing to note is that the input array may have duplicate points. Because two points determine a line, a very intuitive solution is to calculate every two points form a straight line, and then merge the same straight line, the final solution of this problem is to include the number of points on a straight line up to the point. We know that a straight line can be expressed by the slope and y-intercept of the two floating-point (x-axis perpendicular to the slope of the line is infinite, the intercept with the x-intercept), but also need to save the points on each line (to avoid duplication). It sounds very troublesome, but there is a simple way to achieve based on this idea: to a point O as the center, and the slope of the other points to calculate it, and if there are two points A, B and O points form the slope equal, then the ABO three points are collinear, if there are a p
Date : 2025-12-29 Size : 1kb User : lb
[Data structs] Max-Points-on-a-Line DL : 0: 给定的n个点在二维平面上,发现的最大数量点躺在同一直线。-Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
Date : 2025-12-29 Size : 1kb User : 胡延旭
[Data structs] Dijkstra DL : 0: 最小点对问题（二维）二维最接近点对问题：给定平面上n个点，找其中的一对点，使得在n个点的所有点对中，该点对的距离最小。严格地说，最接近点对可能多于1对。【本算法基于C++语言编写，在Windows平台的DEV C++下编译通过，且运行正常】 -The minimum point of the problem (two-dimensional) two-dimensional closest point of the problem: a pair of points given n points in the plane, looking for them, so that all points of n points, the point of minimum distance. Strictly speaking, the closest point to the possibility of more than one pair. The algorithm is based on C++ written language.
Date : 2025-12-29 Size : 2kb User : LEE