ACM POJ 1328 Radar Installation

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld
& %llu

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.



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

题目大意： x轴上方是海，下面是陆地，海面上给出若干个小岛，现在给出灯塔可见半径d和每个灯塔的坐标，要求在陆地上选择建最少的灯塔保证每个岛都能可见，无解输出-1；

分析：首先如果无解，那么一定存在某个灯塔的坐标y大于灯塔半径k，直接输出-1；

如果有解，那么每个灯塔在海岸线上一定存在一个灯塔的可建范围l,r;

此时我们只需要从这些区间当中找到最少灯塔即可。

那么问题来了，如何找呢？

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贪心！

我们将每个小岛按照右范围r从小到大排好，从最左边的岛开始，只要接下来的岛的左范围l小于当前岛的r，那么这两个岛一定可以共用一个灯塔，一直到范围r外的小岛，那么此时肯定需要建立其他灯塔才能满足解。

代码如下：

<script src="https://code.csdn.net/snippets/741681.js"></script>