UVa 10136 - Chocolate Chip Cookies

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题目：平面上有很多个点，画一个直径的5cm的圆，问这个圆最多能包含几个点。

分析：计算几何。一直半径和圆周上两点可以确定2个（或1个）圆，枚举圆周上的两端，统计即可。

时间复杂度为O（n^3），如果两点距离大于直径直接不用计算。

说明：第600题了╮(╯▽╰)╭。

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#include <cmath>using namespace std;typedef struct pnode{double x,y,r;}point;point P[202],cl,cr;double dist_p2p(point a, point b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}//一直圆上两点和半径求圆心 void circle(point a, point b, double r, point &c1, point &c2){double dist_c2ab = sqrt(r*r-dist_p2p(a, b)*dist_p2p(a, b)/4);if (a.x == b.x) {c1.y = c2.y = (a.y+b.y)/2;c1.x = a.x + dist_c2ab;c2.x = a.x - dist_c2ab;}else {double k = atan((b.x-a.x)/(a.y-b.y));c1.x = (a.x+b.x)/2+dist_c2ab*cos(k);c1.y = (a.y+b.y)/2+dist_c2ab*sin(k);c2.x = (a.x+b.x)/2-dist_c2ab*cos(k);c2.y = (a.y+b.y)/2-dist_c2ab*sin(k);}}int main(){int  n;char buf[100];scanf("%d",&n);getchar();getchar();while (n --) {int count = 0;while (gets(buf) && buf[0]) {sscanf(buf,"%lf%lf",&P[count].x,&P[count].y);count ++;}int max_count = 1;for (int i = 1; i < count; ++ i)for (int j = 0; j < i; ++ j)if (dist_p2p(P[i], P[j]) <= 5) {circle(P[i], P[j], 2.5, cl, cr);int l_count = 0,r_count = 0;for (int k = 0; k < count; ++ k) {if (dist_p2p(P[k], cl) <= 2.5)l_count ++;if (dist_p2p(P[k], cr) <= 2.5)r_count ++;}if (max_count < l_count)max_count = l_count;if (max_count < r_count)max_count = r_count;}printf("%d\n",max_count);if (n) printf("\n");}    return 0;}

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