poj 1328 -- Radar Installation

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 49381		Accepted: 11033

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


这个题可以用贪心做，我们以每个岛作为圆心，以r为半径画圆，会交到横坐标上两点（l,r），将两点存起来然后排序，可以按l排，也可以按r来排，这个随意，如果没有交点，则输出-1.
排完序之后，会发现有很多重叠的区域，找到有几个重叠区域即可。

/*======================================================================
 *           Author :   kevin
 *         Filename :   RadarInstallation.cpp
 *       Creat time :   2014-05-14 15:46
 *      Description :
 ========================================================================*/
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#define clr(a,b) memset(a,b,sizeof(a))
#define M 1005
#define INF (1<<30)
using namespace std;

struct Node
{
	double x,y;
	double l,r;
}node[M];
int n,m;
bool cmp(struct Node a,struct Node b)
{
	if(a.r == b.r){
		return a.l < b.l;
	}
	return a.r>b.r;
}
void calcu(int i)
{
	double r = sqrt(m*m-node[i].y*node[i].y);
	node[i].l = node[i].x-r;
	node[i].r = node[i].x+r;
}
int main()
{
	int cas = 1;
	while(scanf("%d%d",&n,&m)!=EOF && n+m){
		clr(node,0);
		int flag = 0;
		for(int i = 0; i < n; i++){
			scanf("%lf%lf",&node[i].x,&node[i].y);
			if(node[i].y > m){
				flag = 1;
			}
			calcu(i);
		}
		if(flag){
			printf("Case %d: -1\n",cas++);
			continue;
		}
		sort(node,node+n,cmp);
		int cnt = 1;
		int s = 0;
		for(int i = 1; i < n; i++){
			if(node[i].r < node[s].l){
				cnt++;
				s = i;
			}
			else if(node[i].l >= node[s].l){
				s = i;
			}
		}
		printf("Case %d: %d\n",cas++,cnt);
	}
	return 0;
}