雷达是怎么计算的，Radar Installation

Question

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

Answer 1

#include
#include
#include
using namespace std;

struct Node {
double left,right;
} str[1005];

bool cmp(Node a,Node b) {
if(a.right==b.right ) {
return a.left>b.left;
} else {
return a.right }
}
int main() {
int n,r,x,y;
int Case=0;
while(~scanf("%d %d",&n,&r)&&n&&r) {
for(int i=0; i scanf("%d %d",&x,&y);
if(y str[i].left=x-sqrt(r*r-y*y);
str[i].right=x+sqrt(r*r-y*y);
} else {
printf("Case %d: -1\n",++Case);
}
}
sort(str,str+n,cmp);
double index=str[0].right;
int num=1;
for(int i=1; i if(str[i].left>index) {
index=str[i].right;
num++;
}
}
printf("Case %d: %d\n",++Case,num);
}
return 0;
}