UVALive-2519 Radar installation Problem solving experience

Original question:

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 Be covered by a RADIUS installation, if the distance between them are 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.

Input

The input consists of several test cases. The first line of each case contains integers n (1n) 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 nlines each containing and integers representing the coordinate of the position of EAC H 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

Test instructions

Given an axis, the positive half axis of y is the sea, the negative half axis is the continent, the X axis is the coastline;

Then there are many islands on the sea that require radar surveillance, and now the question is, how many radars are needed to cover all these islands at least?

Analysis: Can be resolved into the region to select the least point problem, pay attention to replace OK

Code:

#include <iostream>#include<iomanip>#include<cstdio>#include<cmath>#include<algorithm>#include<cstring>using namespacestd;structnode{Doublex, y; BOOL operator< (node &b)Const    {        returnY<b.y | | (y = = b.y&&x>b.x); }}point[ ++Ten];intMain () {intN, D; Doublex, y; intKase =0;  while(CIN >> n >> d&&n! =0&& d! =0) {Kase++; intOK =1; intCNT =0;  for(inti =0; I < n; i++) {scanf ("%LF%LF", &x, &y); if(y>d) {OK=0; }            Else{point[i].x= X-sqrt (d*d-y*y); Point[i].y= x + sqrt (d*d-y*y);        }} getchar (); Sort (point, Point+N); Doubleend;  for(inti =0; I < n; i++)        {            if(i = =0) {End=point[i].y; Continue; }            if(End <point[i].x) {End=point[i].y; CNT++; }        }        if(OK = =0) {printf ("Case %d: -1\n", Kase); }        Else{printf ("Case %d:%d\n", Kase, CNT +1); }    }        return 0;}

UVALive-2519 Radar installation problem solving experience