Radar Installation(贪心）

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 56826		Accepted: 12814

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 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1

 #include<stdio.h>
 #include<iostream>
 #include<algorithm>
 #include<math.h>
 using namespace std;
 struct island
 {
     double x , y ;
     double left , right ;
 }a[];
 int n ;
 int d ;
 bool cmp (island a , island b)
 {
     return a.x < b.x ;
 }

 int main ()
 {
   //  freopen ("a.txt" , "r" , stdin) ;
     int cnt ;
     int ans =  ;
     bool flag ;

     while (~ scanf ("%d%d" , &n , &d)) {
             if (n ==  && d == )
                 break ;
             flag =  ;

         for (int i =  ; i < n ; i++) {
             scanf ("%lf%lf" , &a[i].x , &a[i].y) ;
             if (a[i].y > 1.0 * d || a[i].y <  || d <= ) {
                 flag =  ;
             }
             if (!flag) {
                 a[i].left = (double) a[i].x - sqrt (1.0 * d * d - a[i].y * a[i].y) ;
                 a[i].right = (double) a[i].x + sqrt (1.0 * d * d - a[i].y * a[i].y) ;
             }
         }
         if (flag) {
             printf ("Case %d: -1\n" , ans++) ;
             continue ;
         }
         sort (a , a + n , cmp) ;
         cnt =  ;
         double l = a[].left , r = a[].right ;
         for (int i =  ; i < n ; i++) {
                     if (a[i].left > r) {
                             cnt++ ;
                             l = a[i].left ;
                             r = a[i].right ;
                         }
                     else {
                         l = a[i].left ;
                         r = a[i].right < r ? a[i].right : r ;
                     }
                 }
         printf ("Case %d: %d\n" , ans++ , cnt) ;
     }
     return  ;
 }

别忘记每次都要跟新放radar的区间 ，orz