POJ 1328 Radar Installation

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 75258		Accepted: 16841

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, Def Ining 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 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 each test case output one line consisting of the test case number followed by the minimal number of radar Insta Llations needed. "-1" installation means no solution for this 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

Source Beijing 2002

The problem is that the ship to the island's range, coincident, no matter, there is no overlap in the addition of a radar.

The solution is inverted, and the method has a record value.

#include <stdio.h> #include <math.h> #include <string.h> #include <algorithm> using namespace
Std
int n,d; struct point {double x, y;}
A[1111]; struct DIS {double s,e;}
B[1111];
	BOOL cmp (dis X,dis y) {return x.e<y.e;} double com1 (double x,double y) {double sum=sqrt (1.0*d*d-y*y);
return x-sum;
	} double com2 (double x,double y) {double sum=sqrt (1.0*d*d-y*y);
return x+sum;
	} int main () {double s,e,temp;
	int i,count=1; while (scanf ("%d%d", &n,&d), n| |
		d) {int flag=0;
			for (i=0;i<n;i++) {scanf ("%lf%lf", &a[i].x,&a[i].y);
				if (a[i].y>d) {flag=1;
			Continue } s=com1 (A[I].X,A[I].Y);
			E=com2 (A[I].X,A[I].Y);
				if (s>e) {temp=s;
				S=e;
			E=temp;
			} b[i].s=s;
		b[i].e=e;
		} if (flag) printf ("Case%d: -1\n", count++);
			else {sort (b,b+n,cmp);
			Double max=b[0].e;
			int Ans=1;
				for (i=1;i<n;i++) {if (MAX&GT;=B[I].S) continue;
				MAX=B[I].E;
			ans++; } printf ("CASE%d:%d\n ", Count++,ans);
}} return 0; }