hdoj Radar Installation

[align=left]Problem Description[/align]
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

 

[align=left]Input[/align]
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

 

[align=left]Output[/align]
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.
 

[align=left]Sample Input[/align]

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

 

[align=left]Sample Output[/align]

Case 1: 2
Case 2: 1

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
struct node
{
float l,r,x,y;
}d[1100];
int cmp(node x,node y)
{
return x.l<y.l;/*按照区域的最左端排序*/
}
int main()
{
int sum,t,m,n,flog=1,Case=1,i,num,j;
while(scanf("%d%d",&m,&n),m||n)
{
for(i=0;i<m;i++)
{
scanf("%f%f",&d[i].x,&d[i].y);
if(d[i].y>n)
{
flog=0;break;
}
else
{
d[i].l=d[i].x-sqrt(n*n-d[i].y*d[i].y);/*求出雷达的扫描区域*/
d[i].r=d[i].x+sqrt(n*n-d[i].y*d[i].y);
}
}
printf("Case %d: ",Case);
if(flog==0) printf("-1\n");
else
{
int sign=0;num=0;
sort(d,d+m,cmp);
num++;sign=d[0].r;/*这个很重要，sign始终标记雷达所能扫描的最右端*/
for(i=1;i<m;i++)/*至少设置一个雷达，如果当前判断的区域达不到雷达最右端，加设一个雷达*/
{
if(d[i].r<sign)
sign=d[i].r;
else if(d[i].l>sign)
{
num++;sign=d[i].r;
}
}
printf("%d\n",num);
}
}
return 0;
}