中南大学算法设计与分析实验报告.doc

实验一 分治 —最近点对 问题 Problem Have you ever played quoit in a playground? Quoit is a game in which flat rings are pitched at some toys, with all the toys encircled awarded.In the field of Cyberground, the position of each toy is fixed, and the ring is carefully designed so it can only encircle one toy at a time. On the other hand, to make the game look more attractive, the ring is designed to have the largest radius. Given a configuration of the field, you are supposed to find the radius of such a ring.Assume that all the toys are points on a plane. A point is encircled by the ring if the distance between the point and the center of the ring is strictly less than the radius of the ring. If two toys are placed at the same point, the radius of the ring is considered to be 0. Input The input consists of several test cases. For each case, the first line contains an integer N (2 = N = 100,000), the total number of toys in the field. Then N lines follow, each contains a pair of (x, y) which are the coordinates of a toy. The input is terminated by N = 0. Output For each test case, print in one line the radius of the ring required by the Cyberground manager, accurate up to 2 decimal places. 分析思路 题目是给n个点的坐标，求距离最近的一对点之间距离的一半。第一行是一个数n表示有n个点，接下来n行是n个点的x坐标和y坐标。 首先，假设点是n个，编号为1到n。找一个中间的编号mid，先求出1到mid点的最近距离设为d1，还有mid+1到n的最近距离设为d2。如果说最近点对中的两点都在1-mid集合中，或者mid+1到n集合中，则d就是最小距离了。但是还有可能的是最近点对中的两点分属这两个集合，若存在，则把这个最近点对的距离记录下来，去更新d。这样就得到最小的距离d了。 源代码 #includestdio.h #includemath.h #includealgorithm using namespace std; #define N 1000010 struct point { double x; double y; }p1[N],p2[N]; double dis ( point a , point b ) { return sqrt( pow (a.x-b.x,2) + pow ( a.y-b.y,2 ) ); } double min ( double a , double b ) { return ab?a:b; } bool cpx ( point a , point b ) { return a.x b.x ; } bool cpy ( point a , point b ) { return a.y b.y ; } double mindis ( int l , int r ) { if( l + 1 == r ) return dis ( p1[l] ,p1[r] ); if( l + 2 == r ) return min ( dis ( p1[l