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(三角剖分)三角形和矩形的面积交 2016湖南省赛

  1 // (三角剖分)三角形和矩形的面积交 2016湖南省赛  2   3 #include <iostream>  4 #include <cstdio>  5 #include <cstring>  6 #include <stack>  7 #include <queue>  8 #include <cmath>  9 #include <algorithm> 10 using namespace std; 11  12 const int maxn = 555; 13 const int maxisn = 10; 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16  17 int dcmp(double x) { 18     if(x > eps) return 1; 19     return x < -eps ? -1 : 0; 20 } 21  22 struct Point { 23     double x, y; 24     Point() { 25         x = y = 0; 26     } 27     Point(double a, double b) { 28         x = a, y = b; 29     } 30     inline Point operator-(const Point &b)const { 31         return Point(x - b.x, y - b.y); 32     } 33     inline Point operator+(const Point &b)const { 34         return Point(x + b.x, y + b.y); 35     } 36     inline double dot(const Point &b)const { 37         return x * b.x + y * b.y; 38     } 39     inline double cross(const Point &b, const Point &c)const { 40         return (b.x - x) * (c.y - y) - (c.x - x) * (b.y - y); 41     } 42 }; 43  44 Point LineCross(const Point &a, const Point &b, const Point &c, const Point &d) { 45     double u = a.cross(b, c), v = b.cross(a, d); 46     return Point((c.x * v + d.x * u) / (u + v), (c.y * v + d.y * u) / (u + v)); 47 } 48  49 double PolygonArea(Point p[], int n) { 50     if(n < 3) return 0.0; 51     double s = p[0].y * (p[n - 1].x - p[1].x); 52     p[n] = p[0]; 53     for(int i = 1; i < n; ++ i) 54         s += p[i].y * (p[i - 1].x - p[i + 1].x); 55     return fabs(s * 0.5); 56 } 57  58 double CPIA(Point a[], Point b[], int na, int nb) { //ConvexPolygonIntersectArea 59     Point p[maxisn], tmp[maxisn]; 60     int i, j, tn, sflag, eflag; 61     a[na] = a[0], b[nb] = b[0]; 62     memcpy(p, b, sizeof(Point) * (nb + 1)); 63     for(i = 0; i < na && nb > 2; ++ i) { 64         sflag = dcmp(a[i].cross(a[i + 1], p[0])); 65         for(j = tn = 0; j < nb; ++ j, sflag = eflag) { 66             if(sflag >= 0) tmp[tn ++] = p[j]; 67             eflag = dcmp(a[i].cross(a[i + 1], p[j + 1])); 68             if((sflag ^ eflag) == -2) 69                 tmp[tn ++] = LineCross(a[i], a[i + 1], p[j], p[j + 1]); 70         } 71         memcpy(p, tmp, sizeof(Point) * tn); 72         nb = tn, p[nb] = p[0]; 73     } 74     if(nb < 3) return 0.0; 75     return PolygonArea(p, nb); 76 } 77  78 double SPIA(Point a[], Point b[], int na, int nb) { //SimplePolygonIntersectArea 79     int i, j; 80     Point t1[4], t2[4]; 81     double res = 0, if_clock_t1, if_clock_t2; 82     a[na] = t1[0] = a[0], b[nb] = t2[0] = b[0]; 83     for(i = 2; i < na; ++ i) { 84         t1[1] = a[i - 1], t1[2] = a[i]; 85         if_clock_t1 = dcmp(t1[0].cross(t1[1], t1[2])); 86         if(if_clock_t1 < 0) std::swap(t1[1], t1[2]); 87         for(j = 2; j < nb; ++ j) { 88             t2[1] = b[j - 1], t2[2] = b[j]; 89             if_clock_t2 = dcmp(t2[0].cross(t2[1], t2[2])); 90             if(if_clock_t2 < 0) std::swap(t2[1], t2[2]); 91             res += CPIA(t1, t2, 3, 3) * if_clock_t1 * if_clock_t2; 92         } 93     } 94     return fabs(res); 95 } 96  97 Point p1[maxn], p2[maxn]; 98 int n1, n2; 99 100 int main() {101     double x1,x2,x3,x4,y1,y2,y3,y4;102     while(~scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2)){103         scanf("%lf%lf%lf%lf",&x3,&y3,&x4,&y4);104         p1[0].x=x1,p1[0].y=y1;105         p1[1].x=x1,p1[1].y=y2;106         p1[2].x=x2,p1[2].y=y1;107         p2[0].x=x3,p2[0].y=y3;108         p2[1].x=x3,p2[1].y=y4;109         p2[2].x=x4,p2[2].y=y4;110         p2[3].x=x4,p2[3].y=y3;111         double ans=SPIA(p1,p2,3,4);112         printf("%.6f\n",ans);113     }114     return 0;115 }

 

(三角剖分)三角形和矩形的面积交 2016湖南省赛