HDU - 1007 Quoit Design

最新推荐文章于 2022-02-01 15:29:41 发布

qwerty559

最新推荐文章于 2022-02-01 15:29:41 发布

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本文链接：https://blog.youkuaiyun.com/qq_42367703/article/details/90814815

本文介绍了一个游戏场中计算最优圆环半径的问题，通过枚举并优化算法找到使得圆环能恰好覆盖一个玩具且具有最大半径的解决方案。文章详细描述了输入输出格式，以及使用C++实现的具体算法。

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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.

Sample Input

Sample Output

0.71
0.00
0.75

分析：这个实际上就是一个最近点对问题。暴力枚举再优化一下。

#include<stdio.h>
#include<algorithm>
#include<math.h>
using namespace std;

const int maxn = 100005;

typedef struct Point{
	double x, y;
}Point;

Point p[maxn];

bool cmp(Point a, Point b)
{
	if (a.x == b.x) return a.y < b.y;
	return a.x < b.x;
}

double dis(Point a, Point b)
{
	return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

int main()
{
	int n;
	while (scanf("%d", &n) && n) {
		for (int i = 0; i < n; ++i) {
			scanf("%lf%lf", &p[i].x, &p[i].y);
		}
		sort(p, p + n, cmp);
		double ans = dis(p[0], p[1]), f, ff;
		for (int i = 0; i < n - 1; ++i) {
			for (int j = i + 1; j < n; ++j) {
				if (j == i + 1) ff = f;
				f = dis(p[i], p[j]);
				if (f < ans) ans = f;
				if (f > ff) break;
				ff = f;
			}
		}
		printf("%.2lf\n", ans / 2);
	}
	return 0;
}