LA 2402 (枚举) Fishnet

题意：

正方形四个边界上分别有n个点，将其划分为(n+1)²个四边形，求四边形面积的最大值。

分析：

因为n的规模很小，所以可以二重循环枚举求最大值。

求直线(a, 0) (b, 0) 和直线(0, c) (0, d)的交点，我是二元方程组求解得来的，然后再用叉积求面积即可。

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

const int maxn = 30 + 10;
struct HEHE
{
    double a, b, c, d;
}hehe[maxn];

struct Point
{
    double x, y;
    Point(double x=0, double y=0):x(x), y(y) {}
};
typedef Point Vector;

Vector operator - (const Vector& A, const Vector& B)
{ return Vector(A.x - B.x, A.y - B.y); }

double Cross(const Vector& A, const Vector& B)
{ return (A.x*B.y - A.y*B.x); }

Point GetIntersection(const double& a, const double& b, const double& c, const double& d)
{
    double x = (a+(b-a)*c) / (1-(b-a)*(d-c));
    double y = (d-c)*x+c;
    return Point(x, y);
}

int main(void)
{
    //freopen("2402in.txt", "r", stdin);
    
    int n;
    while(scanf("%d", &n) == 1 && n)
    {
        memset(hehe, 0, sizeof(hehe));
        for(int i = 1; i <= n; ++i) scanf("%lf", &hehe[i].a);
        for(int i = 1; i <= n; ++i) scanf("%lf", &hehe[i].b);
        for(int i = 1; i <= n; ++i) scanf("%lf", &hehe[i].c);
        for(int i = 1; i <= n; ++i) scanf("%lf", &hehe[i].d);
        hehe[n+1].a = hehe[n+1].b = hehe[n+1].c = hehe[n+1].d = 1.0;
        
        double ans = 0.0;
        for(int i = 0; i <= n; ++i)
            for(int j = 0; j <= n; ++j)
            {
                Point A, B, C, D;
                A = GetIntersection(hehe[i].a, hehe[i].b, hehe[j].c, hehe[j].d);
                B = GetIntersection(hehe[i+1].a, hehe[i+1].b, hehe[j].c, hehe[j].d);
                C = GetIntersection(hehe[i+1].a, hehe[i+1].b, hehe[j+1].c, hehe[j+1].d);
                D = GetIntersection(hehe[i].a, hehe[i].b, hehe[j+1].c, hehe[j+1].d);
                double temp = 0.0;
                temp += Cross(B-A, C-A) / 2;
                temp += Cross(C-A, D-A) / 2;
                ans = std::max(ans, temp);
            }
            
        printf("%.6f
", ans);
    }
    
    return 0;
}