【POJ】1328 - Radar Installation（贪心）

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Radar Installation

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

Source

Beijing 2002

首先对于每一个岛，可以用 x ± sqrt （d * d - y * y ）求出的这个区间表示可以覆盖这个岛的雷达的位置。当然如果 y > d 的话这个岛就没法覆盖，就直接输出 -1 结束了。

然后对于这么些区间，如果有某些区间是有重叠的，那么这些岛放置一个雷达就可以全部覆盖，现在问题就是怎么解决区间覆盖了。

我们把雷达的结束位置排一下序（sort），然后从头开始，雷达肯定是放在结束的位置才最有利于后面的岛来 “ 蹭 ” 雷达。然后用一个数组记录一下哪些岛已经被雷达信号覆盖，对于已经被覆盖的，可以直接跳过。

代码如下：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
struct node
{
	double st,endd;
}radar[1011];
bool cmp(node a , node b)
{
	return a.endd < b.endd;
}
int main()
{
	int n,d;
	int Case = 1;
	bool ans;
	bool used[1011];
	int ant;
	while (~scanf ("%d %d",&n,&d) && (n || d))
	{
		ans = true;
		for (int i = 0 ; i < n ; i++)
		{
			double x,y;
			scanf ("%lf %lf",&x,&y);
			if (y > d)
			{
				ans = false;
				continue;
			}
			double t = sqrt(d*d - y*y);
			radar[i].st = x - t;
			radar[i].endd = x + t;
		}
		printf ("Case %d: ",Case++);
		if (!ans)
		{
			printf ("-1\n");
			continue;
		}
		ant = 0;
		sort(radar , radar + n , cmp);		//在结束点安装，所以按结束位置排序 
		memset (used , false , sizeof(used));
		for (int i = 0 ; i < n ; i++)
		{
			if (used[i])
				continue;
			ant++;
			for (int j = i + 1 ; j < n ; j++)
			{
				if (used[j])
					continue;
				if (radar[j].st <= radar[i].endd)
					used[j] = true;
			}
		}
		printf ("%d\n",ant);
	}
	return 0;
}