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
先算出每个岛放探测雷达的x坐标范围，b为雷达探测半径，岛的坐标为（x，y），雷达的坐标范围是（x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)），定义sum=1，i从1开始，每个坐标范围左右边界与上一个坐标范围的右边界比较，具体看代码。
1 #include<cstdio> 2 #include<algorithm> 3 #include<cmath> 4 using namespace std; 5 struct stu 6 { 7 double l,r; 8 } st[1500]; 9 bool cmp(stu a,stu b) 10 { 11 return a.l<b.l; 12 } 13 int main() 14 { 15 int a,s,k=0,sum,i; 16 double b,x,y,sky; 17 while(scanf("%d %lf",&a,&b) &&!(a==0&&b==0)) 18 { 19 s=0; 20 k++; 21 for(i = 0 ; i < a ; i++) 22 { 23 scanf("%lf %lf",&x,&y); 24 if(b < 0 || y > b) 25 { 26 s=-1; 27 } 28 st[i].l=x-sqrt(b*b-y*y); 29 st[i].r=x+sqrt(b*b-y*y); 30 } 31 if(s == -1) printf("Case %d: -1 ",k); 32 else 33 { 34 sort(st,st+a,cmp); 35 sum=1; 36 sky=st[0].r; 37 for(i = 1 ; i < a ; i++) 38 { 39 if(st[i].r <= sky) 40 { 41 sky=st[i].r; 42 } 43 else 44 { 45 if(st[i].l > sky) 46 { 47 sky=st[i].r; 48 sum++; 49 } 50 } 51 } 52 printf("Case %d: %d ",k,sum); 53 } 54 55 } 56 }