UVAlive 2519 Radar Installation (区间选点问题)

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.

Input

The input consists of several test cases. The first line of each case contains two integers n (1n1000)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

题意：给定n个岛屿坐标，和雷达半径，雷达只能放在x轴上，求出最少放几个雷达。

思路：贪心。每个岛屿都有最左和最右最远放雷达能覆盖到的点，我们把这作为左右区间。只要在区间中选中一个位置放雷达。就可以满足该岛屿被覆盖，转换为区间选点问题。

代码：

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

double dd;
int n, i, judge, num, ans, j;
struct D {
	double x;
	double y;
	double l;
	double r;
	int v;
} d[1005];

int cmp(D a, D b) {
	if (a.r != b.r)
		return a.r < b.r;
	return a.l > b.l;
}
int main() {
	int t = 1;
	while (~scanf("%d%lf", &n, &dd) && n || dd) {
		judge = 1; num = 0; ans = 0;
		memset(d, 0, sizeof(d));
		for (i = 0; i < n; i ++) {
			scanf("%lf%lf", &d[i].x, &d[i].y);
			if (d[i].y > dd)
				judge = 0;
			d[i].r = sqrt(dd * dd - d[i].y * d[i].y) + d[i].x;
			d[i].l = d[i].x - sqrt(dd * dd - d[i].y * d[i].y);

		}
		sort(d, d + n, cmp);
		printf("Case %d: ", t ++);
		if (judge) {
			while (num < n) {
				for (i = 0; i < n; i ++) {
					if (!d[i].v) {
						double x = d[i].r;
						for (j = i; j < n; j ++) {
							if (d[j].l <= x && !d[j].v) {
								d[j].v = 1;
								num ++;
							}
						}
						ans ++;
						break;
					}
				}
			}
			printf("%d\n", ans);
		}
		else printf("-1\n");
	}
	return 0;
}