POJ1328 Radar Installation 解题报告

Description

Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d.
We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates.

Figure A Sample Input of Radar Installations

Input

The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases.
The input is terminated by a line containing pair of zeros

Output

For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. "-1" installation means no solution for that case.

Sample Input

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1

分析：

简单的贪心，先按x轴排序，记录圆心位置范围，如果一个点的圆心范围和前面一个圆心范围有交集，就把前一个圆心范围更新为他们的交集

不然答案+1，将当前圆心范围记录下来，最后输出ans

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
using namespace std;
struct node {
	int x, y;
};
node g[1111];
double posr[1111], posl[1111];
int n, d, ans;
int cmp (node a, node b) {
	if (a.x == b.x) return a.y > b.y;
	return a.x < b.x;
}
double fx (node x) {
	double k = sqrt (1.*d * d - x.y * x.y);
	return k;
}
int main() {
	for (int t = 1; cin >> n >> d; t++) {
		if (n == 0 && d == 0) break;
		ans = 0;
		for (int i = 1; i <= n; i++) {
			cin >> g[i].x >> g[i].y;
			if (g[i].y > d) ans = -1;
		}
		if (ans == 0) {
			sort (g + 1, g + 1 + n, cmp);
			if (n >= 1) {
				posl[++ans] = g[1].x - fx (g[1]);
				posr[ans] = g[1].x + fx (g[1]);
			}
			for (int i = 2; i <= n; i++) {
				if (g[i].x == g[i - 1].x) continue;
				if (g[i].x - fx (g[i]) > posr[ans]) {
					posl[++ans] = g[i].x - fx (g[i]), posr[ans] = g[i].x + fx (g[i]);
					continue;
				}
				posl[ans] = max (posl[ans], g[i].x - fx (g[i]) ), posr[ans] = min (posr[ans], g[i].x + fx (g[i]) );
			}
		}
		printf ("Case %d: %d\n", t, ans);
	}
	return 0;
}

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