POJ 1328 Radar Installation greedy algorithm

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 1

1 20 20 0

Sample Output

Case 1:2case 2:1

Source

#include <cstdio>#include<Set>#include<map>#include<cstring>#include<algorithm>#include<queue>#include<iostream>#include<string>#include<cmath>using namespaceStd;typedefLong LongLL;#defineMAXN 1003/*for each point, you can draw a range of centers on the axis that can cover that point, converting to many segments on the x-axis to select a certain number of points, so that all segments contain at least one point sorted by the ending endpoint, and then try to distribute the radar to the right end (so that it overlaps as many segments as possible)*/structnode{DoubleBeg,end;} A[MAXN];intn,d;voidCalDoubleXDoubleYDouble&beg,Double&end)//Request Y<=d{    DoubleR = (Double) D; Beg= X-sqrt (r*r-y*y); End= x + sqrt (r*r-y*y);}BOOLCMP (node A,node b) {returna.end<b.end;}intMain () {intCAS =1;  while(SCANF ("%d%d", &n,&d), n+d) {Doublex, y; BOOLf =false;  for(intI=0; i<n;i++) {scanf ("%LF%LF",&x,&y); if(!f&&y<=D) Cal (X,y,a[i].beg,a[i].end); Else{f=true; }        }        if(f) {printf ("Case %d: -1\n", cas++); Continue; } sort (A,a+n,cmp); intCNT =1; DoubleTMP = a[0].end;  for(intI=1; i<n;i++)        {            if(a[i].beg<=tmp)//because it's sorted by the end,//so the a[i].end must be greater than or equal to TMP, which means no new radars need to be added                Continue; Else//New endpoint start cannot be included, then reset a radar (set at the right end of the new segment){cnt++; TMP=A[i].end; }} printf ("Case %d:%d\n", cas++, CNT); }    return 0;}

POJ 1328 Radar Installation greedy algorithm