Radar Installation
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 60120 | Accepted: 13552 |
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
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
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#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <cctype>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
struct node//记录每个岛的安装雷达的范围
{
double L;
double R;
}point[1100];
bool cmp(node a,node b)//sort比较函数
{
return a.L<b.L;
}
int main()
{
int n,d;
int x,y;
int w=1;
bool flag;
while(scanf("%d %d",&n,&d))
{
if(n==0&&d==0)
{
break;
}
int top=0;
flag=true;
for(int i=0;i<n;i++)
{
scanf("%d %d",&x,&y);
if(d<y)//如果有不符合的记录
flag=false;
if(flag)
{
point[top].L=x-sqrt(d*d-y*y);
point[top].R=x+sqrt(d*d-y*y);
top++;
}
}
printf("Case %d: ",w++);
if(flag)
{
sort(point,point+top,cmp);
double ans=point[0].R;
int sum=1;
for(int i=1;i<top;i++)
{
if(point[i].L>ans)//如果按装的范围不能包括，就增加雷达的个数
{
ans=point[i].R;
sum++;
}
else if(point[i].R<ans)
{
ans=point[i].R;//雷达范围的更新
}
}
printf("%d\n",sum);
}
else
{
printf("-1\n");
}
}
return 0;
}
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