uva 10245 The Closest Pair Problem

题目大意：给出n个点，求出距离最近的两点间的距离。若点与点间的距离都大于10000，输出INFINITY

解题思路：这题的正统做法是分治，偷懒方法是暴力加剪枝。先按x坐标排序，然后fabs(p[i] - p[j]) > ans的点就可以直接跳过了。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<math.h>
using namespace std;
struct point{
	double x, y;
};
double cal(double x1, double y1, double x2, double y2) {
	return sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1));
}
point p[10005];
int cmp(point a, point b) {
		return a.x < b.x;
}
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 = 0xFFFFFFF;
		for (int i = 0; i < n; i++) {
			for (int j = i + 1; j < n; j++) {
				if (fabs(p[i].x - p[j].x) - ans > 1e-9)	break;
				double temp = cal(p[i].x, p[i].y, p[j].x, p[j].y);
				if (temp - ans < 1e-9) {
					ans = temp;
				}
			}
		}
		if (ans - 10000 < 0) {
			printf("%.4lf\n", ans);
		}
		else printf("INFINITY\n");
	}
	return 0;
}