1328 Radar Installation

Radar Installation
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 20456 Accepted: 4160
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 above 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
Source
Beijing 2002

#include<iostream>#include<cmath>using namespace std;struct coordinate//坐标结构体(x,y){int x,y;}p[1001];struct interval//区间结构体[a,b]{double a;double b;}b[1001];int cmp_interval(const void *a,const void *b){interval *A = (interval*) a;interval *B = (interval*) b;return (*A).b > (*B).b ? 1: -1;}//qsort的比较函数double search(coordinate a,int d){double x;x = sqrt((double)(d*d - a.y*a.y)) + a.x;return x;}//找出所给坐标对应的圆心允许范围int main(){int n,d,c = 0,r;//r = number of radar,c = casebool impossible;while(cin >> n >> d){if(n == 0)break;++c;impossible = 0;r = 1;for(int i = 0;i < n;++i){cin >> p[i].x >> p[i].y;if(abs(p[i].y) > d)impossible = 1;b[i].a = 2*p[i].x - search(p[i],d);b[i].b = search(p[i],d);}qsort(b,n,sizeof(b[0]),cmp_interval);//将区间按右边界从小到大排序if(impossible)cout <<"Case "<< c <<": " << -1 <<endl;else{int temp = 0;//贪心过程，策略是对包含区间取最右端的点，若左边界大于之前区间的右边界，则点数必须增加才能满足条件for(int i = 0;i < n;++i){if(b[i].a > b[temp].b){++r;temp = i;}}cout <<"Case "<< c <<": " << r << endl;}}return 0;}