计算几何板子题【2019牛客国庆集训派对day7——三角形和矩形】【多边形相交的面积】

链接：https://ac.nowcoder.com/acm/contest/1112/J
来源：牛客网

题目描述

Bobo 有一个三角形和一个矩形，他想求他们交的面积。
具体地，三角形和矩形由 8 个整数 x1,y1,x2,y2,x3,y3,x4,y4x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4x1​,y1​,x2​,y2​,x3​,y3​,x4​,y4​ 描述。
表示三角形的顶点坐标是 (x1,y1),(x1,y2),(x2,y1)(x_1, y_1), (x_1, y_2), (x_2, y_1)(x1​,y1​),(x1​,y2​),(x2​,y1​)，
矩形的顶点坐标是 (x3,y3),(x3,y4),(x4,y4),(x4,y3)(x_3, y_3), (x_3, y_4), (x_4, y_4), (x_4, y_3)(x3​,y3​),(x3​,y4​),(x4​,y4​),(x4​,y3​).

输入描述:

输入包含不超过 30000 组数据。
每组数据的第一行包含 4 个整数 x1,y1,x2,y2x_1, y_1, x_2, y_2x1​,y1​,x2​,y2​ (x1≠x2,y1≠y2x_1 \neq x_2, y_1 \neq y_2x1​​=x2​,y1​​=y2​).
第二行包含 4 个整数 x3,y3,x4,y4x_3, y_3, x_4, y_4x3​,y3​,x4​,y4​ (x3<x4,y3<y4x_3 < x_4, y_3 < y_4x3​<x4​,y3​<y4​).
(0≤xi,yi≤1040 \leq x_i, y_i \leq 10^40≤xi​,yi​≤104)

输出描述:

对于每组数据，输出一个实数表示交的面积。绝对误差或相对误差小于 10−610^{-6}10−6 即认为正确。

示例1

输入

复制

1 1 3 3
0 0 2 2

输出

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1.00000000

示例2

输入

复制

0 3 3 1
0 0 2 2

输出

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0.75000000

示例3

输入

复制

4462 1420 2060 2969
4159 257 8787 2970

输出

复制

439744.13967527
AC代码：

 /*
  * 多边形的交，多边形的边一定是要按逆时针方向给出
  * 还要判断是凸包还是凹包，调用相应的函数
  * 面积并，只要和面积减去交即可
  */
 #include <bits/stdc++.h>
 using namespace std;  

 const int maxn = ;
 const double eps = 1e-;
 int dcmp(double x)
 {
     if(x > eps) return ;
     return x < -eps ? - : ;
 }
 struct Point
 {
     double x, y;
 };
 double cross(Point a,Point b,Point c) ///叉积
 {
     return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);
 }
 Point intersection(Point a,Point b,Point c,Point d)
 {
     Point p = a;
     double t =((a.x-c.x)*(c.y-d.y)-(a.y-c.y)*(c.x-d.x))/((a.x-b.x)*(c.y-d.y)-(a.y-b.y)*(c.x-d.x));
     p.x +=(b.x-a.x)*t;
     p.y +=(b.y-a.y)*t;
     return p;
 }
 //计算多边形面积
 double PolygonArea(Point p[], int n)
 {
     if(n < ) return 0.0;
     double s = p[].y * (p[n - ].x - p[].x);
     p[n] = p[];
     for(int i = ; i < n; ++ i)
         s += p[i].y * (p[i - ].x - p[i + ].x);
     return fabs(s * 0.5);
 }
 double CPIA(Point a[], Point b[], int na, int nb)//ConvexPolygonIntersectArea
 {
     Point p[], tmp[];
     int tn, sflag, eflag;
     a[na] = a[], b[nb] = b[];
     memcpy(p,b,sizeof(Point)*(nb + ));
     for(int i = ; i < na && nb > ; i++)
     {
         sflag = dcmp(cross(a[i + ], p[],a[i]));
         for(int j = tn = ; j < nb; j++, sflag = eflag)
         {
             if(sflag>=) tmp[tn++] = p[j];
             eflag = dcmp(cross(a[i + ], p[j + ],a[i]));
             if((sflag ^ eflag) == -)
                 tmp[tn++] = intersection(a[i], a[i + ], p[j], p[j + ]); ///求交点
         }
         memcpy(p, tmp, sizeof(Point) * tn);
         nb = tn, p[nb] = p[];
     }
     if(nb < ) return 0.0;
     return PolygonArea(p, nb);
 }
 double SPIA(Point a[], Point b[], int na, int nb)///SimplePolygonIntersectArea 调用此函数
 {
     int i, j;
     Point t1[], t2[];
     double res = , num1, num2;
     a[na] = t1[] = a[], b[nb] = t2[] = b[];
     for(i = ; i < na; i++)
     {
         t1[] = a[i-], t1[] = a[i];
         num1 = dcmp(cross(t1[], t1[],t1[]));
         if(num1 < ) swap(t1[], t1[]);
         for(j = ; j < nb; j++)
         {
             t2[] = b[j - ], t2[] = b[j];
             num2 = dcmp(cross(t2[], t2[],t2[]));
             if(num2 < ) swap(t2[], t2[]);
             res += CPIA(t1, t2, , ) * num1 * num2;
         }
     }
    // res=fabs(res);
     return res;
 }
 Point p1[maxn], p2[maxn];
 int n1, n2;
 int main()
 {
            int x1,x2,x3,x4,y1,y2,y3,y4;
        while(~    scanf("%d %d %d %d",&x1,&y1,&x2,&y2)){

            scanf("%d %d %d %d",&x3,&y3,&x4,&y4);
               p1[].x = x1;p1[].y = y1;
         p1[].x = x1;p1[].y = y2;
         p1[].x = x2;p1[].y = y1;

         p2[].x = x3;p2[].y = y3;
         p2[].x = x3;p2[].y = y4;
         p2[].x = x4;p2[].y = y4;
         p2[].x = x4;p2[].y = y3;
          double s1 = SPIA(p1,p2,,);
         printf("%.8f\n",fabs(s1));
     }
     return ;
 }