POJ1981-Circle and Points（已知两点坐标求圆心坐标）

Circle and Points

Time Limit: 5000MS Memory Limit: 30000KTotal Submissions: 8100 Accepted: 2890Case Time Limit: 2000MS

Description

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

36.47634 7.696285.16828 4.799156.69533 6.2037867.15296 4.083286.50827 2.694665.91219 3.866615.29853 4.160976.10838 3.460396.34060 2.4159987.90650 4.017464.10998 4.183544.67289 4.018876.33885 4.283884.98106 3.827285.12379 5.164737.84664 4.676934.02776 3.87990206.65128 5.474906.42743 6.261896.35864 4.616116.59020 4.542284.43967 5.700594.38226 5.705365.50755 6.181637.41971 6.136686.71936 3.044965.61832 4.238575.99424 4.293285.60961 4.329986.82242 5.796835.44693 3.827246.70906 3.657367.89087 5.680006.23300 4.595305.92401 4.923296.24168 3.813896.22671 3.622100

Sample Output

25511

Source

Japan 2004 Domestic

题意：给你n个点的坐标，求出单位圆能覆盖的最多的点

解题思路：暴力枚举两个在圆上的点，然后在判断其它点能不能落在圆内

#include <iostream>  #include <cstdio>  #include <cstring>  #include <string>  #include <algorithm>  #include <cmath>  #include <map>  #include <set>  #include <stack>  #include <queue>  #include <vector>  #include <bitset>  #include <functional>  using namespace std;#define LL long long  const int INF = 0x3f3f3f3f;struct point{double x, y;}p[400];double dis(point a, point b){return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));}void get(point p1, point p2, point ¢er){point k;k.x = (p1.x + p2.x) / 2.0; k.y = (p1.y + p2.y) / 2.0;double angle = atan2(p1.x - p2.x, p2.y - p1.y);double dcm = sqrt(1 - dis(p1, k)*dis(p1, k));center.x = k.x + dcm*cos(angle); center.y = k.y + dcm*sin(angle);}int main(){int n;while (~scanf("%d", &n) && n){for (int i = 1; i <= n; i++) scanf("%lf%lf", &p[i].x, &p[i].y);int ma = 1;for (int i = 1; i <= n; i++){for (int j = i; j <= n; j++){if (i == j || dis(p[i], p[j])>2.0) continue;int sum = 2;point center;get(p[i], p[j], center);for (int k = 1; k <= n; k++){if (k == i || k == j) continue;if (dis(p[k], center)<1.0) sum++;}ma = max(ma, sum);}}printf("%d\n", ma);}return 0;}