Description
Assume the coasting is a infinite straight line. Side of coasting, sea in the other. Each of small island was a point locating in the sea side. and any radar installation, locating on the coasting, can only cover D-distance, so a island in the sea can is 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 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 a 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 all case contains-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 was followed by n lines each containing and 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 all test case output one line consisting of the test case number followed by the minimal number of radar installation S needed. "-1" installation means no solution for this case.
Sample Input
3 21 2-3 12 11 20 20 0
Sample Output
Case 1:2case 2:1
#include <cstdio> #include <cstring> #include <cmath> #include <iostream> #include < Algorithm>using namespace Std;struct island{double left;double right;} Island[1100];bool CMP (island A,island B) {return a.left<b.left;} int main () {int n,d,i,j,t,temp=0;double x,y,fuck;while (~scanf ("%d%d", &n,&d), n+d) {int m=0;temp++;for (i=0;i <n;i++) {scanf ("%lf%lf", &x,&y); if (y>d| | d<0) {m=1;} island[i].left=x-(sqrt (d*d-y*y)); island[i].right=x+ (sqrt (d*d-y*y));} if (m) {printf ("Case%d: -1\n", temp); continue;} Sort (island,island+n,cmp); T=1;fuck=island[0].right;for (i=1;i<n;i++) {if (Island[i].right<=fuck) {fuck= Island[i].right;} else if (Island[i].left>fuck) {t++;fuck=island[i].right;}} printf ("Case%d:%d\n", temp,t);} return 0;}
POJ-1328 Radar installation-Greedy algorithm