不需要正规的三角剖分，用求多边形面积的思想，从一点出发连接多边形的边得到很多三
角形，三角形有向边方向决定有向面积有正有负，相加得到多边形面积的正值或负值。
把两个多边形都分成若干这样的三角形，求每对三角形的交，根据两三角形有向边顺逆时
针关系确定相交面积的正负号，最后两多边形面积和减去相交面积。
1 #include<stdio.h> 2 #include<string.h> 3 #include<stdlib.h> 4 #include<math.h> 5 #include<algorithm> 6 const int maxn = 555; 7 const int maxisn = 10; 8 const double eps = 1e-8; 9 const double pi = acos(-1.0); 10 int dcmp(double x) 11 { 12 if(x > eps) return 1; 13 return x < -eps ? -1 : 0; 14 } 15 inline double min(double a, double b) 16 {return a < b ? a : b;} 17 inline double max(double a, double b) 18 {return a > b ? a : b;} 19 inline double Sqr(double x) 20 {return x * x;} 21 struct Point 22 { 23 double x, y; 24 Point(){x = y = 0;} 25 Point(double a, double b) 26 {x = a, y = b;} 27 inline Point operator-(const Point &b)const 28 {return Point(x - b.x, y - b.y);} 29 inline Point operator+(const Point &b)const 30 {return Point(x + b.x, y + b.y);} 31 inline double dot(const Point &b)const 32 {return x * b.x + y * b.y;} 33 inline double cross(const Point &b, const Point &c)const 34 {return (b.x - x) * (c.y - y) - (c.x - x) * (b.y - y);} 35 }; 36 Point LineCross(const Point &a, const Point &b, const Point &c, const Point &d) 37 { 38 double u = a.cross(b, c), v = b.cross(a, d); 39 return Point((c.x * v + d.x * u) / (u + v), (c.y * v + d.y * u) / (u + v)); 40 } 41 double PolygonArea(Point p[], int n) 42 { 43 if(n < 3) return 0.0; 44 double s = p[0].y * (p[n - 1].x - p[1].x); 45 p[n] = p[0]; 46 for(int i = 1; i < n; ++ i) 47 s += p[i].y * (p[i - 1].x - p[i + 1].x); 48 return fabs(s * 0.5); 49 } 50 double CPIA(Point a[], Point b[], int na, int nb)//ConvexPolygonIntersectArea 51 { 52 Point p[maxisn], tmp[maxisn]; 53 int i, j, tn, sflag, eflag; 54 a[na] = a[0], b[nb] = b[0]; 55 memcpy(p, b, sizeof(Point) * (nb + 1)); 56 for(i = 0; i < na && nb > 2; ++ i) 57 { 58 sflag = dcmp(a[i].cross(a[i + 1], p[0])); 59 for(j = tn = 0; j < nb; ++ j, sflag = eflag) 60 { 61 if(sflag >= 0) tmp[tn ++] = p[j]; 62 eflag = dcmp(a[i].cross(a[i + 1], p[j + 1])); 63 if((sflag ^ eflag) == -2) 64 tmp[tn ++] = LineCross(a[i], a[i + 1], p[j], p[j + 1]); 65 } 66 memcpy(p, tmp, sizeof(Point) * tn); 67 nb = tn, p[nb] = p[0]; 68 } 69 if(nb < 3) return 0.0; 70 return PolygonArea(p, nb); 71 } 72 double SPIA(Point a[], Point b[], int na, int nb)//SimplePolygonIntersectArea 73 { 74 int i, j; 75 Point t1[4], t2[4]; 76 double res = 0, if_clock_t1, if_clock_t2; 77 a[na] = t1[0] = a[0], b[nb] = t2[0] = b[0]; 78 for(i = 2; i < na; ++ i) 79 { 80 t1[1] = a[i - 1], t1[2] = a[i]; 81 if_clock_t1 = dcmp(t1[0].cross(t1[1], t1[2])); 82 if(if_clock_t1 < 0) std::swap(t1[1], t1[2]); 83 for(j = 2; j < nb; ++ j) 84 { 85 t2[1] = b[j - 1], t2[2] = b[j]; 86 if_clock_t2 = dcmp(t2[0].cross(t2[1], t2[2])); 87 if(if_clock_t2 < 0) std::swap(t2[1], t2[2]); 88 res += CPIA(t1, t2, 3, 3) * if_clock_t1 * if_clock_t2; 89 } 90 } 91 return PolygonArea(a, na) + PolygonArea(b, nb) - res; 92 } 93 Point p1[maxn], p2[maxn]; 94 int n1, n2; 95 int main() 96 { 97 int i; 98 while(scanf("%d%d", &n1, &n2) != EOF) 99 { 100 for(i = 0; i < n1; ++ i) scanf("%lf%lf", &p1[i].x, &p1[i].y); 101 for(i = 0; i < n2; ++ i) scanf("%lf%lf", &p2[i].x, &p2[i].y); 102 printf("%.2f\n", SPIA(p1, p2, n1, n2) + eps); 103 } 104 return 0; 105 }