反正每一个岛都要有雷达覆盖,放雷达的时候尽可能多覆盖岛......
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
struct node
{
double x,y;
};
bool cmp(node a,node b)
{
return a.x<b.x;
}
int main()
{
// freopen("in","r",stdin);
int T,i,n,ans,jishu=0;
double d,pr,l,r,t;
node box[1010];
while(cin>>n>>d)
{
if(n==0&&d==0)
break;
for(i=0;i<n;i++)
cin>>box[i].x>>box[i].y;
sort(box,box+n,cmp);
t=sqrt((d+box[0].y)*(d-box[0].y));
pr=box[0].x+t;
ans=1;
for(i=0;i<n;i++)
{
if(d<fabs(box[i].y))
{
ans=-1;
break;
}
t=sqrt((d+box[i].y)*(d-box[i].y));
l=box[i].x-t;
r=box[i].x+t;
if(l>pr)
{
pr=r;
ans++;
}
else if(r<pr)
pr=r;
}
printf("Case %d: %d\n",++jishu,ans);
}
}
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.
Input
The input consists of several test cases. The first line of each case contains two integers
n (1
n1000)
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