UVa 1301 - Fishnet

求出所有交点枚举每个四边形找最大面积即可。

 #include <cstdio>
 #include <cmath>
 #include <algorithm>

 using namespace std;

 const int MAXN = ;

 const double eps = 1e-;

 struct Point
 {
     double x, y;
     Point( double x = , double y =  ):x(x), y(y) { }
 };

 typedef Point Vector;

 Vector operator+( Vector A, Vector B )       //向量加
 {
     return Vector( A.x + B.x, A.y + B.y );
 }

 Vector operator-( Vector A, Vector B )       //向量减
 {
     return Vector( A.x - B.x, A.y - B.y );
 }

 Vector operator*( Vector A, double p )      //向量数乘
 {
     return Vector( A.x * p, A.y * p );
 }

 Vector operator/( Vector A, double p )      //向量数除
 {
     return Vector( A.x / p, A.y / p );
 }

 bool operator<( const Point& A, const Point& B )   //两点比较
 {
     return A.x < B.x || ( A.x == B.x && A.y < B.y );
 }

 int dcmp( double x )    //控制精度
 {
     if ( fabs(x) < eps ) return ;
     else return x <  ? - : ;
 }

 double Dot( Vector A, Vector B )    //向量点乘
 {
     return A.x * B.x + A.y * B.y;
 }

 double Length( Vector A )           //向量模
 {
     return sqrt( Dot( A, A ) );
 }

 double Angle( Vector A, Vector B )    //向量夹角
 {
     return acos( Dot(A, B) / Length(A) / Length(B) );
 }

 double Cross( Vector A, Vector B )   //向量叉积
 {
     return A.x * B.y - A.y * B.x;
 }

 double Area2( Point A, Point B, Point C )    //向量有向面积
 {
     return Cross( B - A, C - A );
 }

 Vector Rotate( Vector A, double rad )    //向量旋转
 {
     return Vector( A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad) );
 }

 Point GetLineIntersection( Point P, Vector v, Point Q, Vector w )   //两直线交点
 {
     Vector u = P - Q;
     double t = Cross( w, u ) / Cross( v, w );
     return P + v * t;
 }

 bool SegmentProperIntersection( Point a1, Point a2, Point b1, Point b2 )  //线段相交，交点不在端点
 {
     double c1 = Cross( a2 - a1, b1 - a1 ), c2 = Cross( a2 - a1, b2 - a1 ),
                 c3 = Cross( b2 - b1, a1 - b1 ), c4 = Cross( b2 - b1, a2 - b1 );
     return dcmp(c1)*dcmp(c2) <  && dcmp(c3) * dcmp(c4) < ;
 }

 bool OnSegment( Point p, Point a1, Point a2 )   //点在线段上，不包含端点
 {
     return dcmp( Cross(a1 - p, a2 - p) ) ==  && dcmp( Dot( a1 - p, a2 - p ) ) < ;
 }

 Point P[][MAXN];
 Point ch[MAXN][MAXN];

 void init( int n )
 {
     ch[][].x = 0.0, ch[][].y = 0.0;
     ch[][n + ].x = 1.0, ch[][n + ].y = 0.0;
     ch[n + ][].x = 0.0, ch[n + ][].y = 1.0;
     ch[n + ][n + ].x = 1.0, ch[n + ][n + ].y = 1.0;

     for ( int i = ; i <= n; ++i )
         ch[][i] = P[][i];
     for ( int i = ; i <= n; ++i )
         ch[n + ][i] = P[][i];
     for ( int i = ; i <= n; ++i )
         ch[i][] = P[][i];
     for ( int i = ; i <= n; ++i )
         ch[i][n + ] = P[][i];

     for ( int i = ; i <= n; ++i )
         for ( int j = ; j <= n; ++j )
             ch[i][j] = GetLineIntersection( P[][i], P[][i] - P[][i], P[][j], P[][j] - P[][j] );

     return;
 }

 int main()
 {
     int n;
     while ( scanf( "%d", &n ), n )
     {
         for ( int i = ; i < ; ++i )
         {
             for ( int j = ; j <= n; ++j )
             {
                 double a;
                 scanf( "%lf", &a );
                 switch( i )
                 {
                 case :
                     P[i][j].x = a;
                     P[i][j].y = 0.0;
                     break;
                 case :
                     P[i][j].x = a;
                     P[i][j].y = 1.0;
                     break;
                 case :
                     P[i][j].x = 0.0;
                     P[i][j].y = a;
                     break;
                 case :
                     P[i][j].x = 1.0;
                     P[i][j].y = a;
                     break;
                 }
             }
         }

         init( n );

         double ans = 0.0;
         for ( int i = ; i <= n; ++i )
             for ( int j = ; j <= n; ++j )
             {
                 Point a = ch[i][j], b = ch[i][j + ], c = ch[i + ][j + ], d = ch[i + ][j];
                 double area = fabs( Cross( b-a, c-a )/2.0 ) + fabs( Cross( c-a, d-a )/2.0 );
                 if ( area > ans ) ans = area;
             }

         printf( "%.6f\n", ans );

     }
     return ;
 }