POJ-1328Radar Installation（贪心）

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 67259		Accepted: 15091

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

反向考虑，把雷达建立到岛上，则其圆形区域与x轴相交的弦上每个点都可以放雷达侦测到这个岛屿，同理，求其他所有岛屿，按弦的左边坐标排序，把雷达建立到右点上，如果下一段的左端点大于前一个的右端点，则需要一个新的雷达，且雷达建立在这一段弦的右端点，如果下一段的左端点小于前一个的右端点，则不需要新建雷达，但是雷达的位置要改变到下一段的右端点，如此重复！

#include
#include
#include
#include
using namespace std;
struct node 
{
	double l,r;
}s[1100];
bool cmp(node x,node y)
{
	return x.lr)
				flag=1;
		}
		if(flag)
		{
			printf("Case %d: -1\n",++cot);
			continue;
		}
		sort(s,s+m,cmp);
		int t=0;
		for(i=1;is[t].r)
			{
				sum++;
				t=i;
			}
			else if(s[i].r<=s[t].r)
				t=i;
		}
		printf("Case %d: %d\n",++cot,sum);
	}
	return 0;
 }