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

 struct Point
 {
     double x, y;
     Point(double x=, double y=):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) / (-(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) ==  && n)
     {
         memset(hehe, , sizeof(hehe));
         for(int i = ; i <= n; ++i) scanf("%lf", &hehe[i].a);
         for(int i = ; i <= n; ++i) scanf("%lf", &hehe[i].b);
         for(int i = ; i <= n; ++i) scanf("%lf", &hehe[i].c);
         for(int i = ; i <= n; ++i) scanf("%lf", &hehe[i].d);
         hehe[n+].a = hehe[n+].b = hehe[n+].c = hehe[n+].d = 1.0;

         double ans = 0.0;
         for(int i = ; i <= n; ++i)
             for(int j = ; 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+].a, hehe[i+].b, hehe[j].c, hehe[j].d);
                 C = GetIntersection(hehe[i+].a, hehe[i+].b, hehe[j+].c, hehe[j+].d);
                 D = GetIntersection(hehe[i].a, hehe[i].b, hehe[j+].c, hehe[j+].d);
                 double temp = 0.0;
                 temp += Cross(B-A, C-A) / ;
                 temp += Cross(C-A, D-A) / ;
                 ans = std::max(ans, temp);
             }

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

     return ;
 }

代码君