UVa 1301 - Fishnet

求出所有交点枚举每个四边形找最大面积即可。

  1 #include <cstdio>

  2 #include <cmath>

  3 #include <algorithm>

  4 

  5 using namespace std;

  6 

  7 const int MAXN = 40;

  8 

  9 const double eps = 1e-10;

 10 

 11 struct Point

 12 {

 13     double x, y;

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

 15 };

 16 

 17 typedef Point Vector;

 18 

 19 Vector operator+( Vector A, Vector B )       //向量加

 20 {

 21     return Vector( A.x + B.x, A.y + B.y );

 22 }

 23 

 24 Vector operator-( Vector A, Vector B )       //向量减

 25 {

 26     return Vector( A.x - B.x, A.y - B.y );

 27 }

 28 

 29 Vector operator*( Vector A, double p )      //向量数乘

 30 {

 31     return Vector( A.x * p, A.y * p );

 32 }

 33 

 34 Vector operator/( Vector A, double p )      //向量数除

 35 {

 36     return Vector( A.x / p, A.y / p );

 37 }

 38 

 39 bool operator<( const Point& A, const Point& B )   //两点比较

 40 {

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

 42 }

 43 

 44 int dcmp( double x )    //控制精度

 45 {

 46     if ( fabs(x) < eps ) return 0;

 47     else return x < 0 ? -1 : 1;

 48 }

 49 

 50 double Dot( Vector A, Vector B )    //向量点乘

 51 {

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

 53 }

 54 

 55 double Length( Vector A )           //向量模

 56 {

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

 58 }

 59 

 60 double Angle( Vector A, Vector B )    //向量夹角

 61 {

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

 63 }

 64 

 65 double Cross( Vector A, Vector B )   //向量叉积

 66 {

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

 68 }

 69 

 70 double Area2( Point A, Point B, Point C )    //向量有向面积

 71 {

 72     return Cross( B - A, C - A );

 73 }

 74 

 75 Vector Rotate( Vector A, double rad )    //向量旋转

 76 {

 77     return Vector( A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad) );

 78 }

 79 

 80 Point GetLineIntersection( Point P, Vector v, Point Q, Vector w )   //两直线交点

 81 {

 82     Vector u = P - Q;

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

 84     return P + v * t;

 85 }

 86 

 87 bool SegmentProperIntersection( Point a1, Point a2, Point b1, Point b2 )  //线段相交，交点不在端点

 88 {

 89     double c1 = Cross( a2 - a1, b1 - a1 ), c2 = Cross( a2 - a1, b2 - a1 ),

 90                 c3 = Cross( b2 - b1, a1 - b1 ), c4 = Cross( b2 - b1, a2 - b1 );

 91     return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;

 92 }

 93 

 94 bool OnSegment( Point p, Point a1, Point a2 )   //点在线段上，不包含端点

 95 {

 96     return dcmp( Cross(a1 - p, a2 - p) ) == 0 && dcmp( Dot( a1 - p, a2 - p ) ) < 0;

 97 }

 98 

 99 Point P[4][MAXN];

100 Point ch[MAXN][MAXN];

101 

102 void init( int n )

103 {

104     ch[0][0].x = 0.0, ch[0][0].y = 0.0;

105     ch[0][n + 1].x = 1.0, ch[0][n + 1].y = 0.0;

106     ch[n + 1][0].x = 0.0, ch[n + 1][0].y = 1.0;

107     ch[n + 1][n + 1].x = 1.0, ch[n + 1][n + 1].y = 1.0;

108 

109     for ( int i = 1; i <= n; ++i )

110         ch[0][i] = P[0][i];

111     for ( int i = 1; i <= n; ++i )

112         ch[n + 1][i] = P[1][i];

113     for ( int i = 1; i <= n; ++i )

114         ch[i][0] = P[2][i];

115     for ( int i = 1; i <= n; ++i )

116         ch[i][n + 1] = P[3][i];

117 

118     for ( int i = 1; i <= n; ++i )

119         for ( int j = 1; j <= n; ++j )

120             ch[i][j] = GetLineIntersection( P[2][i], P[3][i] - P[2][i], P[0][j], P[1][j] - P[0][j] );

121 

122     return;

123 }

124 

125 int main()

126 {

127     int n;

128     while ( scanf( "%d", &n ), n )

129     {

130         for ( int i = 0; i < 4; ++i )

131         {

132             for ( int j = 1; j <= n; ++j )

133             {

134                 double a;

135                 scanf( "%lf", &a );

136                 switch( i )

137                 {

138                 case 0:

139                     P[i][j].x = a;

140                     P[i][j].y = 0.0;

141                     break;

142                 case 1:

143                     P[i][j].x = a;

144                     P[i][j].y = 1.0;

145                     break;

146                 case 2:

147                     P[i][j].x = 0.0;

148                     P[i][j].y = a;

149                     break;

150                 case 3:

151                     P[i][j].x = 1.0;

152                     P[i][j].y = a;

153                     break;

154                 }

155             }

156         }

157 

158         init( n );

159 

160         double ans = 0.0;

161         for ( int i = 0; i <= n; ++i )

162             for ( int j = 0; j <= n; ++j )

163             {

164                 Point a = ch[i][j], b = ch[i][j + 1], c = ch[i + 1][j + 1], d = ch[i + 1][j];

165                 double area = fabs( Cross( b-a, c-a )/2.0 ) + fabs( Cross( c-a, d-a )/2.0 );

166                 if ( area > ans ) ans = area;

167             }

168 

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

170 

171     }

172     return 0;

173 }