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

Sample Output

Case 1: 2
Case 2: 1

 1 #include<stdio.h>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<math.h>
 5 using namespace std;
 6 struct island
 7 {
 8     double x , y ;
 9     double left , right ;
10 }a[2000];
11 int n ;
12 int d ;
13 bool cmp (island a , island b)
14 {
15     return a.x < b.x ;
16 }
17 
18 int main ()
19 {
20   //  freopen ("a.txt" , "r" , stdin) ;
21     int cnt ;
22     int ans = 1 ;
23     bool flag ;
24 
25     while (~ scanf ("%d%d" , &n , &d)) {
26             if (n == 0 && d == 0)
27                 break ;
28             flag = 0 ;
29 
30         for (int i = 0 ; i < n ; i++) {
31             scanf ("%lf%lf" , &a[i].x , &a[i].y) ;
32             if (a[i].y > 1.0 * d || a[i].y < 0 || d <= 0) {
33                 flag = 1 ;
34             }
35             if (!flag) {
36                 a[i].left = (double) a[i].x - sqrt (1.0 * d * d - a[i].y * a[i].y) ;
37                 a[i].right = (double) a[i].x + sqrt (1.0 * d * d - a[i].y * a[i].y) ;
38             }
39         }
40         if (flag) {
41             printf ("Case %d: -1
" , ans++) ;
42             continue ;
43         }
44         sort (a , a + n , cmp) ;
45         cnt = 1 ;
46         double l = a[0].left , r = a[0].right ;
47         for (int i = 1 ; i < n ; i++) {
48                     if (a[i].left > r) {
49                             cnt++ ;
50                             l = a[i].left ;
51                             r = a[i].right ;
52                         }
53                     else {
54                         l = a[i].left ;
55                         r = a[i].right < r ? a[i].right : r ;
56                     }
57                 }
58         printf ("Case %d: %d
" , ans++ , cnt) ;
59     }
60     return 0 ;
61 }

View Code

别忘记每次都要跟新放radar的区间，orz