Poj1328--radar installation (range selected)

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 64121		Accepted: 14418

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

Source

Beijing 2002 Interval Point (conversion idea):
<1&gt: The left endpoint is sorted (small → large), the left endpoint is the same, (small → large) The right end of the row;
<2&gt: For Num[0], on the right end of the radar, if the next interval left endpoint > now right endpoint, radar shu+1, if the left endpoint < right end point: If the next interval right endpoint < Now right endpoint, update the radar to the next interval right end;

1#include <cmath>2#include <cstdio>3#include <algorithm>4 using namespacestd;5 6 struct Island7 {8     DoubleL, R;9} num[1010];Ten  One BOOLCMP (Island L, Island R) A { -     if(L.L = =R.L) -         returnL.R <R.R; the     returnL.L <R.L; - } -  - intMain () + { -     DoubleR, X, y;intI, M, t=1; +      while(~SCANF ("%d%lf", &m, &R)) A     { at         intFlag =0; -         if(M = =0&& r = =0) -              Break; -          for(i=0; i<m; i++) -         { -scanf"%LF%LF", &x, &y); in             if(Y >R) -{flag =1;Continue; } toNUM[I].L = X-sqrt (r*r-y*y);//conversion to interval problem; +NUM[I].R = x + sqrt (r*r-y*y); -         } the         if(flag) *         { $printf"Case %d: -1\n", t++); Panax Notoginseng             Continue; -         } theSort (num, num+m, CMP); +         Doubletemp = num[0].R;intTotal =1; A          for(i=1; i<m; i++)         the         { +             if(Num[i].l >temp) { -total++; $temp =NUM[I].R; $             } -                  -             if(Num[i].l <=temp) the             { -                 if(NUM[I].R <temp)Wuyitemp =NUM[I].R; the             } -         } Wuprintf"Case %d:%d\n", t++, total);  -     } About     return 0; $}

Poj1328--radar installation (range selected)