c++ 多邊形求交集程式碼
阿新 • • 發佈:2018-12-19
/* * 多邊形的交,多邊形的邊一定是要按逆時針方向給出 * 還要判斷是凸包還是凹包,呼叫相應的函式 * 面積並,只要和麵積減去交即可 */ #include <bits/stdc++.h> using namespace std; const int maxn = //300; const double eps = 1e-8; int dcmp(double x) { if(x > eps) return 1; return x < -eps ? -1 : 0; } struct Point { double x, y; }; double cross(Point a,Point b,Point c) ///叉積 { 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; } //計算多邊形面積 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]); ///求交點 } 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 呼叫此函式 { 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; } Point p1[maxn], p2[maxn]; int n1, n2; int main() { while(cin>>n1>>n2) { for(int i = 0; i < n1; i++) scanf("%lf%lf", &p1[i].x, &p1[i].y); for(int i = 0; i < n2; i++) scanf("%lf%lf", &p2[i].x, &p2[i].y); double Area = SPIA(p1, p2, n1, n2); } return 0; }