UVa 1301

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 = 40;

const double eps = 1e-10;

struct Point
{
    double x, y;
    Point( double x = 0, double y = 0 ):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 0;
    else return x < 0 ? -1 : 1;
}

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) < 0 && dcmp(c3) * dcmp(c4) < 0;
}

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) ) == 0 && dcmp( Dot( a1 - p, a2 - p ) ) < 0;
}

Point P[4][MAXN];
Point ch[MAXN][MAXN];

void init( int n )
{
    ch[0][0].x = 0.0, ch[0][0].y = 0.0;
    ch[0][n + 1].x = 1.0, ch[0][n + 1].y = 0.0;
    ch[n + 1][0].x = 0.0, ch[n + 1][0].y = 1.0;
    ch[n + 1][n + 1].x = 1.0, ch[n + 1][n + 1].y = 1.0;

    for ( int i = 1; i <= n; ++i )
        ch[0][i] = P[0][i];
    for ( int i = 1; i <= n; ++i )
        ch[n + 1][i] = P[1][i];
    for ( int i = 1; i <= n; ++i )
        ch[i][0] = P[2][i];
    for ( int i = 1; i <= n; ++i )
        ch[i][n + 1] = P[3][i];

    for ( int i = 1; i <= n; ++i )
        for ( int j = 1; j <= n; ++j )
            ch[i][j] = GetLineIntersection( P[2][i], P[3][i] - P[2][i], P[0][j], P[1][j] - P[0][j] );

    return;
}

int main()
{
    int n;
    while ( scanf( "%d", &n ), n )
    {
        for ( int i = 0; i < 4; ++i )
        {
            for ( int j = 1; j <= n; ++j )
            {
                double a;
                scanf( "%lf", &a );
                switch( i )
                {
                case 0:
                    P[i][j].x = a;
                    P[i][j].y = 0.0;
                    break;
                case 1:
                    P[i][j].x = a;
                    P[i][j].y = 1.0;
                    break;
                case 2:
                    P[i][j].x = 0.0;
                    P[i][j].y = a;
                    break;
                case 3:
                    P[i][j].x = 1.0;
                    P[i][j].y = a;
                    break;
                }
            }
        }

        init( n );

        double ans = 0.0;
        for ( int i = 0; i <= n; ++i )
            for ( int j = 0; j <= n; ++j )
            {
                Point a = ch[i][j], b = ch[i][j + 1], c = ch[i + 1][j + 1], d = ch[i + 1][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
", ans );

    }
    return 0;
}