UVa 11437 - Triangle Fun

我是求出来所有点的坐标用叉积算的面积……

据说可以证明出来S△PQR = S△ABC/7

  1 #include <cstdio>

  2 #include <cmath>

  3 

  4 struct Point

  5 {

  6     double x, y;

  7     Point( double x = 0, double y = 0 ):x(x), y(y) { }

  8 };

  9 

 10 Point operator+( Point A, Point B )

 11 {

 12     return Point( A.x + B.x, A.y + B.y );

 13 }

 14 

 15 Point operator-( Point A, Point B )

 16 {

 17     return Point( A.x - B.x, A.y - B.y );

 18 }

 19 

 20 Point operator*( Point A, double p )

 21 {

 22     return Point( A.x * p, A.y * p );

 23 }

 24 

 25 Point operator/( Point A, double p )

 26 {

 27     return Point( A.x / p, A.y / p );

 28 }

 29 

 30 bool operator<( const Point& A, const Point& B )

 31 {

 32     return A.x < B.x || ( A.x == B.x && A.y < B.y );

 33 }

 34 

 35 Point Rotate( Point A, double rad )

 36 {

 37     return Point( A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad) );

 38 }

 39 

 40 double Cross( Point A, Point B )

 41 {

 42     return A.x * B.y - A.y * B.x;

 43 }

 44 

 45 Point GetLineIntersection( Point P, Point v, Point Q, Point w )

 46 {

 47     Point u = P - Q;

 48     double t = Cross( w, u ) / Cross( v, w );

 49     return P + v * t;

 50 }

 51 

 52 double Dot( Point A, Point B )

 53 {

 54     return A.x * B.x + A.y * B.y;

 55 }

 56 

 57 double Length( Point A )

 58 {

 59     return sqrt( Dot( A, A ) );

 60 }

 61 

 62 double Angle( Point A, Point B )

 63 {

 64     return acos( Dot(A, B) / Length(A) / Length(B) );

 65 }

 66 

 67 Point read_Point()

 68 {

 69     double x, y;

 70     scanf( "%lf%lf", &x, &y );

 71     return Point( x, y );

 72 }

 73 

 74 Point GetD( Point B, Point C )

 75 {

 76     return ( C - B ) / 3.0 + B;

 77 }

 78 

 79 int main()

 80 {

 81     Point A, B, C, D, E, F;

 82     Point P, Q, R;

 83     int T;

 84     scanf( "%d", &T );

 85     while ( T-- )

 86     {

 87         A = read_Point();

 88         B = read_Point();

 89         C = read_Point();

 90         D = GetD( B, C );

 91         E = GetD( C, A );

 92         F = GetD( A, B );

 93         Point AD = D - A;

 94         Point BE = E - B;

 95         Point CF = F - C;

 96         P = GetLineIntersection( B, BE, A, AD );

 97         Q = GetLineIntersection( C, CF, B, BE );

 98         R = GetLineIntersection( A, AD, C, CF );

 99 

100         printf( "%.0f\n", Cross( Q - P, R - P ) / 2.0 );

101     }

102     return 0;

103 }