ACM Radar Placement Problem

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<tex2html_verbatim_mark> distance , so an island in the sea can be covered by a RADIUS installation, if the distance between them are at most D& Lt;tex2html_verbatim_mark>.

We use Cartesian coordinate system, defining the coasting is the x<tex2html_verbatim_mark>-axis. The sea side is above x<tex2html_verbatim_mark>-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<tex2html_verbatim_mark> y< ; tex2html_verbatim_mark>coordinates.

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Input

The input consists of several test cases. The first line of all case contains integers n<tex2html_verbatim_mark>(1n) < Tex2html_verbatim_mark> and D<tex2html_verbatim_mark>, where n<tex2html_ Verbatim_mark> is the number of islands in the sea and D<tex2html_verbatim_mark> is the distance of Coverage of the radar installation. This was followed by n<tex2html_verbatim_mark> lines each containing and integers representing the CO Ordinate 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 the case.

Sample Input

3 21 2-3 12 11 20 20 0

Sample Output

Case 1:2case 2:1

Problem Solving Ideas:

The main idea of this topic is that there is an axis above the x-axis is the land on the x-axis is the coastline, the Sea has n small islands, now to install radar on the coastline, radar coverage is a circle with a radius of R, please cover all the islets with a minimum of radar; output-1 when unable to overwrite. We can figure out the center of the radar where each island can be covered, that is, a circle with a radius of R for the small island, the intersection of the circle and the x-axis: The left intersection is X-SQRT (r*r-y*y); The right intersection is X+SQRT (r*r-y*y), and according to the left intersection, if the left intersection of I points is on the right side of the current radar, a new radar should be installed to update the radar otherwise if the right intersection point of I is also on the left side of the current radar, the center of the current radar is updated to the right intersection of the point.

Program code:

#include <iostream> #include <algorithm> #include <cmath>using namespace std;const int N = 1001;struct Node{double L,r;} A[n];bool Fun (Node A, Node B) {return A.L<B.L;} int main () {int n,r,i,case=0;double x,y,t;while (scanf ("%d%d", &n, &r) &&n&&r) {int v=1;for (i=0; i<n; i++) {cin>>x>>y;a[i].l = X-sqrt (r*r-y*y); A[I].R = x + sqrt (r*r-y*y); if (Y>r | | r<=0 | | y<0) V=-1;} Sort (A, a+n,fun); t = a[0].r;for (i=1; i<n && v!=-1; i++) if (a[i].l>t) {v++;t = A[I].R;} else if (a[i].r<t) t=a[i].r;printf ("Case%d:%d\n", ++case, V);} return 0;}     

　　

ACM Radar Placement Problem