Radar Installation
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 54756 | Accepted: 12337 |
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
/*思路：该题题意是为了求出能够覆盖所有岛屿的最小雷达数目，
每个小岛对应x轴上的一个区间，在这个区间内的任何一个点放置雷达，
则可以覆盖该小岛，区间范围的计算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)];
问题即转化为已知一定数量的区间，求最少区间可以包含所有的点，
使得每个区间内斗至少存在一个点。每次迭代对于第一个区间，
选择最右边一个点， 因为它可以让较多区间得到满足，选择该点之后，
将得到满足的区间删掉， 进行下一步迭代， 直到所有的点均包含则结束。
不过还要考虑d<0和y>d的情况。
左端点：表示雷达能够覆盖岛左临界的雷达，
右端点：表示雷达能够覆盖岛右临界的雷达。
*/
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
struct st
{
double left,right;
}data[1010];
int cmp(st a,st b)
{
return a.left<b.left;
}
int main()
{
int i,t,kase=1,sum,n,d;
double x,y,ans;
while(scanf("%d%d",&n,&d)&&(n||d))
{
t=0;
for(i=0;i<n;i++)
{
scanf("%lf %lf",&x,&y);
if(y>d||d<0)
{
t=1;
}
data[i].left=x-sqrt(d*d-y*y);
data[i].right=x+sqrt(d*d-y*y);
}
sort(data,data+n,cmp);
if(t==1)
{
printf("Case %d: -1\n",kase);
kase++;
}
else
{
ans=data[0].right;
sum=1;
for(i=1;i<n;i++)
{
if(data[i].right<=ans)//如果右端点在前一雷达覆盖范围，
ans=data[i].right;//则雷达要放在放在右端点。
else if(data[i].left>ans)//如果左端点不在前一雷达覆盖范围内，
{ //则需要增加新雷达，并将雷达放在右端点。
sum++;
ans=data[i].right;
}
}
printf("Case %d: %d\n",kase,sum);
kase++;
}
}
return 0;
}