csu----J(1812): 三角形和矩形

多边形交并补模板运用

J(1812): 三角形和矩形题目

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Description

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,y4 描述。 表示三角形的顶点坐标是 (x1,y1),(x1,y2),(x2,y1)， 矩形的顶点坐标是 (x3,y3),(x3,y4),(x4,y4),(x4,y3).

Input

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,y2 (x1≠x2,y1≠y2).

第二行包含 4 个整数 x3,y3,x4,y4 (x3<x4,y3<y4).

(0≤xi,yi≤104)

Output

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 10-6 即认为正确。

Sample Input

1 1 3 3
0 0 2 2
0 3 3 1
0 0 2 2
4462 1420 2060 2969
4159 257 8787 2970

Sample Output

1.00000000
0.75000000
439744.13967527

Hint

 

/*
 * 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 = 100005;
const double eps = 1e-8;
int dcmp(double x)
{
    if(x > eps) return 1;
    return x < -eps ? -1 : 0;
}
struct Point
{
    double x, y;
}z;
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 < 3) return 0.0;
    double s = p[0].y * (p[n - 1].x - p[1].x);
    p[n] = p[0];
    for(int i = 1; i < n; ++ i)
        s += p[i].y * (p[i - 1].x - p[i + 1].x);
    return fabs(s * 0.5);
}
double CPIA(Point a[], Point b[], int na, int nb)//ConvexPolygonIntersectArea
{
    Point p[20], tmp[20];
    int tn, sflag, eflag;
    a[na] = a[0], b[nb] = b[0];
    memcpy(p,b,sizeof(Point)*(nb + 1));
    for(int i = 0; i < na && nb > 2; i++)
    {
        sflag = dcmp(cross(a[i + 1], p[0],a[i]));
        for(int j = tn = 0; j < nb; j++, sflag = eflag)
        {
            if(sflag>=0) tmp[tn++] = p[j];
            eflag = dcmp(cross(a[i + 1], p[j + 1],a[i]));
            if((sflag ^ eflag) == -2)
                tmp[tn++] = intersection(a[i], a[i + 1], p[j], p[j + 1]); ///求交点
        }
        memcpy(p, tmp, sizeof(Point) * tn);
        nb = tn, p[nb] = p[0];
    }
    if(nb < 3) 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[4], t2[4];
    double res = 0, num1, num2;
    a[na] = t1[0] = a[0], b[nb] = t2[0] = b[0];
    for(i = 2; i < na; i++)
    {
        t1[1] = a[i-1], t1[2] = a[i];
        num1 = dcmp(cross(t1[1], t1[2],t1[0]));
        if(num1 < 0) swap(t1[1], t1[2]);
        for(j = 2; j < nb; j++)
        {
            t2[1] = b[j - 1], t2[2] = b[j];
            num2 = dcmp(cross(t2[1], t2[2],t2[0]));
            if(num2 < 0) swap(t2[1], t2[2]);
            res += CPIA(t1, t2, 3, 3) * num1 * num2;
        }
    }
    return res;
}
bool cmp_node(Point a,Point b)
{
    double ans=(b.x-z.x)*(a.y-z.y)-(b.y-z.y)*(a.x-z.x);
    if(fabs(ans)<eps)
        return sqrt(pow(a.x-z.x,2)+pow(a.y-z.y,2))<sqrt(pow(b.x-z.x,2)+pow(b.y-z.y,2));
    return ans>0;
}
void sort_j(Point p[],int n)
{
    int pos=0;
    for(int i=0;i<n;i++)
    {
        if(p[pos].y>p[i].y)
            pos=i;
        if(p[pos].y==p[pos].y&&p[pos].x>p[i].x)
            pos=i;
    }
    swap(p[pos],p[0]);
    z=p[0];
    sort(p+1,p+n,cmp_node);
}
Point p1[maxn], p2[maxn];
int n1, n2;
int main()
{
    double x1,x2,x3,x4,y1,y2,y3,y4;
    while(~scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4))
    {
        p1[0].x=x2;p1[0].y=y1;
        p1[1].x=x1;p1[1].y=y1;
        p1[2].x=x1;p1[2].y=y2;
        p2[0].x=x4;p2[0].y=y3;
        p2[1].x=x3;p2[1].y=y3;
        p2[2].x=x4;p2[2].y=y4;
        p2[3].x=x3;p2[3].y=y4;
        sort_j(p1,3);
        sort_j(p2,4);
        double Area = SPIA(p1, p2,3,4);
        printf("%.8f\n",Area);
    }
    return 0;
}