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;
}

UVa 10136 - Chocolate Chip Cookies

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原文地址：http://blog.csdn.net/mobius_strip/article/details/44872723