ZOJ 1041 Transmitters

题意：给你一个能绕圆心转动的半径固定的半圆和N个点，问这个半圆最多能覆盖多少个点，半圆边界上的点也算在覆盖范围内。

首先把半圆半径之外的点全部排除，枚举剩余所有点与圆心的连线。判断在同一侧的点有多少个。

之前忽略的边界情况，WA一次。

 

View Code

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <cstdlib>

 4 #include <cmath>

 5 

 6 const int MAXN = 200;

 7 const double EPS = 1e-9;

 8 

 9 struct point

10 {

11     double x, y;

12 };

13 

14 point T;

15 point P[MAXN];

16 double R;

17 int N, cnt;

18 

19 int max( int a, int b )

20 {

21     return a > b ? a : b;

22 }

23 

24 double GetDis( point a, point b )

25 {

26     return sqrt( (a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y-b.y) );

27 }

28 

29 void GetLine( point a, point b, double& K, double &B )

30 {

31     K = ( a.y - b.y ) / ( a.x - b.x );

32     B = a.y - K*a.x;

33     return;

34 }

35 

36 void solved()

37 {

38     int ans = 0;

39     double K, B;

40     for ( int i = 0; i < cnt; ++i )

41     {

42         if ( fabs( P[i].x - T.x ) > EPS )

43         {

44             GetLine( T, P[i], K, B );

45             int count1 = 0;

46             int count0 = 0;

47             //printf( "K=%f B=%lf\n", K, B );

48             for ( int j = 0; j < cnt; ++j )

49             {

50                 //printf("y=%f\n", K * P[j].x - P[j].y + B );

51                 double temp = K * P[j].x - P[j].y + B;

52                 if ( temp > 0 ) ++count1;

53                 else if ( fabs( temp ) < EPS ) ++count0;

54             }

55             //printf("1=%d 0=%d\n", count1, count0 );

56             ans = max( ans, max( cnt - count1, count1 + count0 ) );

57         }

58         else

59         {

60             int count1 = 0;

61             int count0 = 0;

62             for ( int j = 0; j < cnt; ++j )

63             {

64                 if ( P[j].x - T.x > 0 ) ++count1;

65                 else if ( fabs( P[j].x - T.x ) < EPS ) ++count0;

66             }

67             //printf("**1=%d 0=%d\n", count1, count0 );

68             ans = max( ans, max( cnt - count1, count1 + count0 ) );

69         }

70     }

71 

72     printf("%d\n", ans );

73     return;

74 }

75 

76 int main()

77 {

78     while ( ~scanf( "%lf%lf%lf", &T.x, &T.y, &R ) )

79     {

80         if ( R < 0 ) break;

81         scanf( "%d", &N );

82         cnt = 0;

83         for ( int i = 0; i < N; ++i )

84         {

85             scanf( "%lf%lf", &P[cnt].x, &P[cnt].y );

86             double temp = GetDis( T, P[cnt] );

87             if ( temp < R || fabs( temp - R ) < EPS ) ++cnt;

88         }

89         //printf("cnt=%d\n", cnt);

90 

91         solved();

92     }

93     return 0;

94 }