2109: Circle and Points
ResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE20s8192K23159Standard
You are given N points in the xy-plane. You have a circle of radius one and move it on the xy-plane, so as to enclose as many of the points as possible. Find how many points can be simultaneously enclosed at the maximum. A point is considered enclosed by a circle when it is inside or on the circle.
Fig 1. Circle and Points
Input
The input consists of a series of data sets, followed by a single line only containing a single character '0', which indicates the end of the input. Each data set begins with a line containing an integer N, which indicates the number of points in the data set. It is followed by N lines describing the coordinates of the points. Each of the N lines has two decimal fractions X and Y, describing the x- and y-coordinates of a point, respectively. They are given with five digits after the decimal point.
You may assume 1 <= N <= 300, 0.0 <= X <= 10.0, and 0.0 <= Y <= 10.0. No two points are closer than 0.0001. No two points in a data set are approximately at a distance of 2.0. More precisely, for any two points in a data set, the distance d between the two never satisfies 1.9999 <= d <= 2.0001. Finally, no three points in a data set are simultaneously very close to a single circle of radius one. More precisely, let P1, P2, and P3 be any three points in a data set, and d1, d2, and d3 the distances from an arbitrarily selected point in the xy-plane to each of them respectively. Then it never simultaneously holds that 0.9999 <= di <= 1.0001 (i = 1, 2, 3).
Output
For each data set, print a single line containing the maximum number of points in the data set that can be simultaneously enclosed by a circle of radius one. No other characters including leading and trailing spaces should be printed.
Sample Input
3
6.47634 7.69628
5.16828 4.79915
6.69533 6.20378
6
7.15296 4.08328
6.50827 2.69466
5.91219 3.86661
5.29853 4.16097
6.10838 3.46039
6.34060 2.41599
8
7.90650 4.01746
4.10998 4.18354
4.67289 4.01887
6.33885 4.28388
4.98106 3.82728
5.12379 5.16473
7.84664 4.67693
4.02776 3.87990
20
6.65128 5.47490
6.42743 6.26189
6.35864 4.61611
6.59020 4.54228
4.43967 5.70059
4.38226 5.70536
5.50755 6.18163
7.41971 6.13668
6.71936 3.04496
5.61832 4.23857
5.99424 4.29328
5.60961 4.32998
6.82242 5.79683
5.44693 3.82724
6.70906 3.65736
7.89087 5.68000
6.23300 4.59530
5.92401 4.92329
6.24168 3.81389
6.22671 3.62210
0
Sample Output
2
5
5
11
这道题根据题目描述，不难想到一个O(n^3)的算法，即每次以O(n^2)的时间枚举两个点，求出过这两个点的单位圆，然后O(n)时间内统计落在圆内的点的个数，然后全局维护一个最大值。可以证明，一个覆盖最多点的单位圆一定至少经过两个点（当然，点集规模N >=2)
不过这样做效率比较低（某个大神的O(n^3)的算法居然在JOJ上跑了3s多。。比好多自称O(n^2 logn)的算法快多了。。），有一种更好的方法时间复杂度是O(n^2*logn)，是根据以每个点为圆心单位长度为半径的圆形两两相交弧来判断。下面贴的代码是看了
http://i.eol.cn/blog_read.php?topicid=91156之后写的，效率很高，在JOJ 的status上面排第一(1.83s 874K),POJ上跑了900多ms，可以把这个作为求单位圆覆盖的模板。
#include <iostream>
#include <cmath>
#include <algorithm>
#define pi acos(-1.0)
using namespace std;
struct point{double x,y;}p[300+5];
int n;
double dis(const point &a,const point &b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double k(const point &a, const point &b)
{
double ki = atan(fabs((b.y-a.y)/(b.x-a.x)));
if(b.y - a.y > 0)
{if(b.x - a.x < 0) ki = pi - ki;}
else
if(b.x < a.x) ki += pi;
else ki = 2*pi - ki;
return ki;
}
struct angle
{
double ang;
bool is;
}even[1000+10];
bool cmp(const angle &a, const angle &b){return a.ang < b.ang;}
int CircleMaxPoint(double r)
{
int ans = 1, i, j, top;
double dist,thi,delta;
for(i = 0; i < n; i++)
{
top = 0;
for(j = 0; j < n; j++)
{
if(i == j) continue;
dist = dis(p[i],p[j]);
if(dist > 2 * r) continue;
thi = k(p[i],p[j]);
delta = acos(dist / 2.0 / r);
even[top].ang = thi - delta; even[top++].is = true;
even[top].ang = thi + delta; even[top++].is = false;
}
if(top < ans) continue;
sort(even,even+top,cmp);
int coun = 1;
for(j = 0; j < top; j++)
{
if(even[j].is)
coun++;
else
coun--;
if(coun > ans)
ans = coun;
}
}
return ans;
}
int main()
{
int t;
while(scanf("%d", &n) != EOF)
{
if(!n) break;
for(int i = 0; i < n; i++)
scanf("%lf %lf", &p[i].x, &p[i].y);
printf("%d/n", CircleMaxPoint(1.0));
}
}
