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 }