nyoj 287 Radar

Radar

时间限制：1000 ms  |  内存限制：65535 KB

难度：3

描述

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.

 

输入

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

输出

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.

样例输入

3 2

1 2

-3 1

2 1

 

1 2

0 2

 

0 0

样例输出

Case 1: 2

Case 2: 1

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct dy
{
	double left;//**左交点**//
	double right;//**右交点**//
}w[1001];
bool comp(dy a,dy b)//**按左交点坐标从小到大排序**//
{
	if(a.left<b.left) return true;
	return false;
}
int main()
{
	int n,r,x,y,i,count,num=1;
	double len,t;
	while(~scanf("%d %d",&n,&r)&&(n,r))
	{
		memset(w,0,sizeof(w));
		count=1;//**从第一个点开始，所以计数器初值为1**//
		for(i=0;i<=n-1;i++)
		{
			scanf("%d %d",&x,&y);
			len=sqrt(((double)r*r-(double)y*y));//**结合图形，勾股定理**//
			w[i].left=(double)x-len;//**左交点的坐标**//
			w[i].right=(double)x+len;//**右交点的坐标**//
		}
		for(i=0;i<=n-1;i++)
		{
			if(y>r)//**如果不能完全覆盖**//
			{
				printf("Case %d: -1\n",num++);
				break;
			}
		}
		sort(w,w+n,comp);
		t=w[0].right;
		for(i=1;i<=n-1;i++)
		{
			if(w[i].left>t)//**如果后一个点的左交点大于前一个点的右坐标，说明两点没有公共区域**//
			{
				count++;
				t=w[i].right;
			}
			else
			{
				if(w[i].right<t)//**如果后一个点的右交点小于前一个点的右坐标，说明后一个点的覆盖区域被前一个点包含了**//
				{
					t=w[i].right;//**保证后一个点被覆盖**//
				}
			}
		}
		printf("Case %d: %d\n",num++,count);
	}
	return 0;
}