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
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\n" , 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\n" , ans++ , cnt) ;
59 }
60 return 0 ;
61 }
别忘记每次都要跟新放radar的区间 ，orz