Poj1328 radar installation (Greedy)

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Question link: http://poj.org/problem? Id = 1328

Description

Assume the coasting is an infinite straight line. land is in one side of coasting, sea in the other. each small island is a point locating in the sea side. and any radar installation, locating on the coasting, can only cover D distance, so an island in the sea can be 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 abve X-axis, and the land side below. given the position of each island in 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 an island 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 each case contains two 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 is followed by n lines each containing two 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 installations needed. "-1" Installation means no solution for that case.

Sample Input

3 21 2-3 12 11 20 20 0

Sample output

Case 1: 2Case 2: 1

Idea: Reverse Thinking: record the left and right ranges that each island can reach with struct, and use greedy ideas to calculate the number of radars;

The Code is as follows:

# Include <iostream> # include <algorithm> # include <cmath> using namespace STD; struct point {Double R, L;} p [1017]; bool CMP (point, point B) {return. L <B. l ;}int main () {int n, m, R, flag; int I, j; int X, Y; int CAS = 0, cont; while (CIN> N> r) {If (n = 0 & R = 0) break; flag = cont = 0; for (I = 0; I <n; I ++) {CIN> x> Y; If (FABS (y)> r) {flag = 1;} p [I]. L = x * 1.000-sqrt (R * r * 1.00-y * 1.000); P [I]. R = x * 1.000 + SQRT (R * r * 1.00-y * 1.000);} cout <"case" <++ CAS <":"; if (FLAG) {cout <"-1" <Endl; continue;} Sort (p, p + N, CMP); cont = 1; double T = P [0]. r; for (I = 1; I <n; I ++) {If (P [I]. r <t) // to prevent the right section of the previous Island from being greater than the right section of the next island {// if the right section of the island is on the left of the radar, it requires the location of the Mobile Radar to the right of the island T = P [I]. r;} If (P [I]. l> T) // If the left area of the next island is on the right side of the radar {// you need to have another radar located on this island. t = P [I]. r; cont ++ ;}}cout <cont <Endl ;}return 0 ;}