Problem D
Subway
Input: standard input
Output: standard output
Time Limit: 4 seconds

The government in a foreign land is looking into the possibility of establishing a subway system in its capital. Because of practical reasons, they would like each subway line to start at the central station and then go in a straight line in some angle as far as necessary. You have been hired to investigate whether such an approach is feasible. You have been given the coordinates of important places in the city which must lie close to a subway station (possibly the central station). Exactly how close they must lie is determined by another parameter, also given. Your job is to calculate the minimum number of subway lines needed to satisfy the demands of the government. You may assume that any number of subway stations can be built along a subway line.

Input

The first line in the input file contains an integer N, the number of data sets to follow (at most 20). Each set starts with two integers, n and d (1<=n<=200, 0<=d<150). n is the number of important places in the city that must have a subway station nearby, and d is the maximum distance allowed between an important place and a subway station. Then comes n lines, each line containing two integers x and y (-100<=x, y<=100), the coordinates of an important place in the capital. The central station will always have coordinates 0,0. All pair of coordinates within a data set will be distinct (and none will be 0,0).

/*The figure above corresponds to the first sample input*/

Output

For each data set, output a single integer on a line by itself: the minimum number of subway lines needed to make sure all important places in the city is at a distance of at most d from a subway station.

Sample Input                                Output for Sample Input  

2

7 1

-1 -4

-3 1

-3 -1

2 3

2 4

2 -2

6 -2

4 0

0 4

-12 18

0 27

-34 51

4

2

题意：要建铁路，以(0,0)为点，给n个站点，要求站点距离不能铁路超过d。问最少建几条铁路。

思路：每个站点可以构造出最左和最右满足的铁路，就形成一个区间，那么就变成区间选点问题，不过由于是一个圆周，要从每个位置作为起点进行一次区间选点。还要注意精度误差。还要注意一开始角度的处理。。

代码：

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#define min(a,b) ((a)<(b)?(a):(b))
using namespace std;
const int N = 205;
const double pi = acos(-1.0);
int t, n, qn;
double d, x, y;
struct Q {
	double l, r;
}q[N * 2];

double cal(double x, double y) {
	return sqrt(x * x + y * y);
}

bool cmp(Q a, Q b) {
	return a.r < b.r;
}

int solve() {
	int ans = qn, i;
	sort(q, q + qn, cmp);
	for (i = 0; i < qn; i++) {
		q[i + qn].l = q[i].l + 2 * pi;
		q[i + qn].r = q[i].r + 2 * pi;
	}
	for (i = 0; i < qn; i++) {
		int num = 1; double v = q[i].r;
		for (int j = i + 1; j < i + qn; j++) {
			if (v - q[j].l < -1e-9) {
				num++;
				v = q[j].r;
			}
		}
		ans = min(ans, num);
	}
	return ans;
}

int main() {
	scanf("%d", &t);
	while (t--) {
		qn = 0;
		scanf("%d%lf", &n, &d);
		for (int i = 0; i < n; i++) {
			scanf("%lf%lf", &x, &y);
			if (cal(x, y) > d) {
				double p = atan2(y, x);
				double du = asin(d / cal(x, y));
				q[qn].l = p - du; q[qn++].r = p + du;
				if (q[qn - 1].r > pi) {
					q[qn - 1].l -= 2 * pi;
					q[qn - 1].r -= 2 * pi;
				}
			}
		}
		printf("%d\n", solve());
	}
	return 0;
}

原文：http://blog.csdn.net/accelerator_/article/details/19417967