poj1328 Radar Installation 解题报告

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

解题思路：看到题目想到了先把输入的点排序，然后求出每个人的可能圆心的范围，从左到右一直找下去， 只要是重合的，就可以归入一个圆心，但要规划圆心的时候，要注意，当圆心归入后，要对他们的 右范围比较，选少的作为新的右范围。这里面还是数据处理比较重要。  代码如下： 

Code:

1. #include<iostream>
2. #include<cmath>
3. #include<algorithm>
4. usingnamespacestd;
5. constintMax(1005);
6. typedefstructdata
7. {
8. doublex1;
9. doublex2;
10. }coordinate;//定义圆心范围的结构体
12. coordinatepoint[Max];
14. intcompare(coordinatea,coordinateb)
15. {
16. return(a.x1-b.x1)<10e-7;
17. }
19. intmain()
20. {
21. intstep=1;
22. while(1)
23. {
24. intflag=0;
25. intn,d;
26. cin>>n>>d;
27. if(n==0&&d==0)
28. break;
29. inta,b;
30. for(inti=0;i<n;i++)//注意这边i不能从1开始，因为后面用到了sort排序
31. {
32. cin>>a>>b;
33. if(!flag&&(b<=d))
34. {
35. doubletemp=sqrt(double(d*d-b*b));//由点处理出圆心的范围
36. point[i].x1=a-temp;
37. point[i].x2=a+temp;
38. }
39. else
40. flag=1;
42. }
43. if(flag==1)
44. {
45. cout<<"Case"<<step<<":"<<"-1"<<endl;
46. step++;
47. }
48. else
49. {
50. intsum=1;
51. sort(point,point+n,compare);
52. doubletemp=point[0].x2;
53. for(inti=1;i<n;i++)
54. {
55. if(point[i].x1-temp>10e-7)
56. {
57. sum++;
58. temp=point[i].x2;
59. }
60. else
61. {
62. if(point[i].x2-temp<10e-7)//当走到的点的右范围比原来的右范围小，即temp相应的改
63. temp=point[i].x2;
64. }
65. }
66. cout<<"Case"<<step<<":"<<sum<<endl;
67. step++;
68. }
69. }
70. return0;
72. }