POJ 1328

 1  

 2 //坐标精度是int

 3 /*

 4 圆心位于 

 5 */

 6 #include <iostream>

 7 #include <cstdlib>

 8 #include <cstring>

 9 #include <cmath>

10 using namespace std; 

11 

12 const int N = 1005;

13 typedef struct Part 

14 {

15     int a, b;

16 };

17 Part q[N];

18 

19 typedef struct Node 

20 {

21     int left, right;

22 };

23 Node node[N];

24 

25 int cmp(const void *a, const void *b)//只能返回int

26 {

27     Part *c = (Part *)a;

28     Part *d = (Part *)b;

29     if(c->b == d->b)

30         return c->a < d->a;

31     return c->b > d->b;

32 }

33 

34 int main()

35 {

36     int i,j,k;

37     int n, d;//d为半径

38     int num = 1;

39     while(cin>>n>>d, n&&d)

40     {

41         /*

42         d<0时输出无解，却忘了发现d<0后，后面还有N行输入要“干”掉.

43         */

44         if(d<=0)

45         {

46             int a, b;

47             for(i=1; i<=n; i++)

48                 cin>>a>>b;

49             cout<<"Case "<<num++<<": "<<-1<<endl;

50             continue;

51         }

52         int cnt = 1;

53         bool flag;

54         for(i=1; i<=n; i++)

55         {

56             cin>>node[i].left>>node[i].right;

57             if(abs(node[i].right)>d)

58             {

59                 flag = true;

60             }

61         }

62         if(flag)

63         {

64             cout<<"Case "<<num++<<": "<<-1<<endl;

65             continue;

66         }

67         

68         for(i=1; i<=n; i++)

69         {//圆心在下面的区间内那么就可以包括那个点 

70             q[i].a = node[i].left - sqrt(1.0*(d*d - node[i].right*node[i].right));

71             q[i].b = node[i].left + sqrt(1.0*(d*d - node[i].right*node[i].right));        

72         }

73         

74         //下面是区间选点问题 

75         qsort(q+1, n, sizeof(Part), cmp);

76         int end = q[1].b;

77         for(i=2; i<=n; i++)

78         {

79             if(end<q[i].a)

80             {

81                 end = q[i].b;

82                 cnt++;

83             }

84         

85         }

86         cout<<"Case "<<num++<<": "<<cnt<<endl; 

87     }

88     return 0; 

89 }

90 

91 

92     

93