Radar Installation
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 99975 | Accepted: 22202 |
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
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
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 <cstring>
#include <cstdio>
#include <cstdlib>
#include <iomanip>
#include <algorithm>
#include <iomanip>
#include <cmath>
using namespace std;
struct node{
double left,rigth;//我们用来记录每个小岛与x轴相交的左右两点
}num[1005];
bool cmp(node a,node b)
{
return a.left<=b.left;//每个小岛左坐标按照横坐标从小到大排序
}
int main()
{
int n,flag,cnt,i,k=1;
double x,y,d,p;
while(cin>>n>>d&&n&&d)//输入数据
{
flag=1;
for(i=0;i<n;i++)
{
cin>>x>>y;
if(d<y)//如果该点的纵坐标比雷达半径大,雷达无法覆盖
{
flag=0;
continue;
}
num[i].left=x-sqrt(d*d-y*y);//求出以小岛为中心,以雷达半径为半径的圆与x轴相交的左右横坐标
num[i].rigth=x+sqrt(d*d-y*y);
}
if(flag)//如果能够找到能够覆盖点的雷达
{
sort(num,num+n,cmp);//从小到大排序
cnt=1;
p=num[0].rigth;//第一个雷达的横坐标
for(i=1;i<n;i++)
{
if(num[i].left>p)
{
p=num[i].rigth;//如果该点的左坐标比雷达大,那么雷达无法覆盖,则增加雷达,并把该点的右坐标设定为雷达
cnt++;
}else if(num[i].rigth<p)//如果该点能够被雷达覆盖,但是我们要求出最少的雷达数,我们需要使雷达尽可能的多覆盖,所以要把雷达的位置缩小
{
p=num[i].rigth;
}
}
cout<<"Case "<<k++<<": "<<cnt<<endl;
}
else{//雷达不能全部覆盖
cout<<"Case "<<k++<<": "<<"-1"<<endl;
}
//cout<<"Case "<<k++<<": "<<cnt<<endl;
}
return 0;
}