半平面交算法及简单应用
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时间：2015年06月08日
半平面交算法及简单应用
半平面：一条直线把二维平面分成两个平面。
半平面交：在二维几何平面上，给出若干个半平面，求它们的公共部分
半平面交的结果：1.凸多边形（后面会讲解到）2.无界，因为有可能若干半平面没有形成封闭3.直线，线段，点，空（属于特殊情况吧）
算法：1：根据上图可以知道，运用给出的多边形每相邻两点形成一条直线来切割原有多边形，如果多边形上的点i在有向直线的左边或者在直线上即保存起来，否则判断此点的前一个点i-1和后一个点i+1是否在此直线的左边或线上，在的话分别用点i和点i-1构成的直线与此时正在切割的直线相交求出交点，这个交点显然也要算在切割后剩下的多边形里，同理点i和点i+1。原多边形有n条边，每条边都要进行切割，所以时间复杂度为O(n^2)。
2：第二种就是训练指南上面详细讲解的运用双端队列的半平面交算法，时间复杂度为O(nlogn)。仔细阅读代码应该能理解。
代码实现：以poj3130为例，裸的模板。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 #include<cmath> 6 #include<algorithm> 7 #define inf 0x7fffffff 8 #define exp 1e-10 9 #define PI 3.141592654 10 using namespace std; 11 const int maxn=111; 12 struct Point 13 { 14 double x,y; 15 Point (double x=0,double y=0):x(x),y(y){} 16 }an[maxn],bn[maxn],cn[maxn]; 17 ///an:记录最开始的多边形；bn:临时保存新切割的多边形；cn:保存新切割出的多边形 18 typedef Point Vector; 19 Vector operator + (Vector A,Vector B) {return Vector(A.x+B.x , A.y+B.y); } 20 Vector operator - (Vector A,Vector B) {return Vector(A.x-B.x , A.y-B.y); } 21 Vector operator * (Vector A,double p) {return Vector(A.x*p , A.y*p); } 22 Vector operator / (Vector A,double p) {return Vector(A.x/p , A.y/p); } 23 int dcmp(double x) {if (fabs(x)<exp) return 0;return x>0 ? 1 : -1; } 24 double cross(Vector A,Vector B) 25 { 26 return A.x*B.y-B.x*A.y; 27 } 28 29 double A,B,C; 30 int n,m; 31 void getline(Point a,Point b)///获取直线 Ax + By + C = 0 32 { 33 A=b.y-a.y; 34 B=a.x-b.x; 35 C=b.x*a.y-a.x*b.y; 36 } 37 ///getline()函数得到的直线和点a和点b所连直线的交点 38 Point intersect(Point a,Point b) 39 { 40 double u=fabs(A*a.x+B*a.y+C); 41 double v=fabs(A*b.x+B*b.y+C); 42 Point ans; 43 ans.x=(a.x*v+b.x*u)/(u+v); 44 ans.y=(a.y*v+b.y*u)/(u+v); 45 return ans; 46 } 47 void cut()///切割，原多边形的点为顺时针存储 48 { 49 int cnt=0; 50 for (int i=1 ;i<=m ;i++) 51 { 52 if (A*cn[i].x + B*cn[i].y + C>=0) bn[++cnt]=cn[i]; 53 else 54 { 55 if (A*cn[i-1].x + B*cn[i-1].y + C > 0) bn[++cnt]=intersect(cn[i-1],cn[i]); 56 if (A*cn[i+1].x + B*cn[i+1].y + C > 0) bn[++cnt]=intersect(cn[i+1],cn[i]); 57 } 58 } 59 for (int i=1 ;i<=cnt ;i++) cn[i]=bn[i]; 60 cn[0]=bn[cnt]; 61 cn[cnt+1]=bn[1]; 62 m=cnt;///新切割出的多边形的点数 63 } 64 void solve() 65 { 66 for (int i=1 ;i<=n ;i++) cn[i]=an[i]; 67 an[n+1]=an[1]; 68 cn[n+1]=an[1]; 69 cn[0]=an[n]; 70 m=n; 71 for (int i=1 ;i<=n ;i++) 72 { 73 getline(an[i],an[i+1]); 74 cut(); 75 } 76 } 77 int main() 78 { 79 while (scanf("%d",&n)!=EOF && n) 80 { 81 for (int i=1 ;i<=n ;i++) scanf("%lf%lf",&an[i].x,&an[i].y); 82 reverse(an+1,an+n+1); 83 solve(); 84 if (m) puts("1"); 85 else puts("0"); 86 } 87 return 0; 88 }
poj3335，给出两种方法
1.O(n^2)
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 #include<cmath> 6 #include<algorithm> 7 #define inf 0x7fffffff 8 #define exp 1e-10 9 #define PI 3.141592654 10 using namespace std; 11 const int maxn=111; 12 struct Point 13 { 14 double x,y; 15 Point (double x=0,double y=0):x(x),y(y){} 16 }an[maxn],bn[maxn],cn[maxn]; 17 typedef Point Vector ; 18 Vector operator + (Vector A,Vector B) {return Vector(A.x+B.x , A.y+B.y); } 19 Vector operator - (Vector A,Vector B) {return Vector(A.x-B.x , A.y-B.y); } 20 Vector operator * (Vector A,double p) {return Vector(A.x*p , A.y*p); } 21 Vector operator / (Vector A,double p) {return Vector(A.x/p , A.y/p); } 22 int dcmp(double x) {if (fabs(x)<exp) return 0;return x>0 ? 1 : -1; } 23 double cross(Vector A,Vector B) 24 { 25 return A.x*B.y-B.x*A.y; 26 } 27 28 int n,m; 29 double A,B,C; 30 void getline(Point a,Point b) 31 { 32 A=b.y-a.y; 33 B=a.x-b.x; 34 C=b.x*a.y-a.x*b.y; 35 } 36 Point intersect(Point a,Point b) 37 { 38 double u=fabs(A*a.x+B*a.y+C); 39 double v=fabs(A*b.x+B*b.y+C); 40 Point ans; 41 ans.x=(a.x*v+b.x*u)/(u+v); 42 ans.y=(a.y*v+b.y*u)/(u+v); 43 return ans; 44 } 45 void cut() 46 { 47 int cnt=0; 48 for (int i=1 ;i<=m ;i++) 49 { 50 if (A*cn[i].x+B*cn[i].y+C>=0) 51 bn[++cnt]=cn[i]; 52 else 53 { 54 if (A*cn[i-1].x+B*cn[i-1].y+C>0) 55 bn[++cnt]=intersect(cn[i-1],cn[i]); 56 if (A*cn[i+1].x+B*cn[i+1].y+C>0) 57 bn[++cnt]=intersect(cn[i+1],cn[i]); 58 } 59 } 60 for (int i=1 ;i<=cnt ;i++) cn[i]=bn[i]; 61 cn[0]=bn[cnt]; 62 cn[cnt+1]=bn[1]; 63 m=cnt; 64 } 65 void solve() 66 { 67 for (int i=1 ;i<=n ;i++) cn[i]=an[i]; 68 an[n+1]=an[1]; 69 cn[n+1]=cn[1]; 70 cn[0]=cn[n]; 71 m=n; 72 for (int i=1 ;i<=n ;i++) 73 { 74 getline(an[i],an[i+1]); 75 cut(); 76 } 77 } 78 int main() 79 { 80 int t; 81 scanf("%d",&t); 82 while (t--) 83 { 84 scanf("%d",&n); 85 for (int i=1 ;i<=n ;i++) scanf("%lf%lf",&an[i].x,&an[i].y); 86 solve(); 87 if (!m) printf("NO\n"); 88 else printf("YES\n"); 89 } 90 return 0; 91 }
2.O(nlogn)
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 #include<cmath> 6 #include<algorithm> 7 #define inf 0x7fffffff 8 #define exp 1e-10 9 #define PI 3.141592654 10 using namespace std; 11 const int maxn=111; 12 struct Point 13 { 14 double x,y; 15 Point (double x=0,double y=0):x(x),y(y){} 16 }an[maxn]; 17 typedef Point Vector; 18 struct Line 19 { 20 Point p; 21 Vector v; 22 double ang; 23 Line (){} 24 Line (Point p,Vector v):p(p),v(v){ang=atan2(v.y,v.x); } 25 //Line (Point p,Vector v):p(p),v(v) {ang=atan2(v.y,v.x); } 26 friend bool operator < (Line a,Line b) 27 { 28 return a.ang<b.ang; 29 } 30 }bn[maxn]; 31 Vector operator + (Vector A,Vector B) {return Vector(A.x+B.x , A.y+B.y); } 32 Vector operator - (Vector A,Vector B) {return Vector(A.x-B.x , A.y-B.y); } 33 Vector operator * (Vector A,double p) {return Vector(A.x*p , A.y*p); } 34 Vector operator / (Vector A,double p) {return Vector(A.x/p , A.y/p); } 35 int dcmp(double x) {if (fabs(x)<exp) return 0;return x>0 ? 1 : -1; } 36 double cross(Vector A,Vector B) 37 { 38 return A.x*B.y-B.x*A.y; 39 } 40 bool OnLeft(Line L,Point p) 41 { 42 return cross(L.v,p-L.p)>=0;///点P在有向直线L的左边（>=0说明在线上也算） 43 } 44 Point GetIntersection(Line a,Line b) 45 { 46 Vector u=a.p-b.p; 47 double t=cross(b.v,u)/cross(a.v,b.v); 48 return a.p+a.v*t; 49 } 50 //Point GetIntersection(Line l1, Line l2) { 51 // Point p; 52 // double dot1,dot2; 53 // //dot1 = multi(l2.a, l1.b, l1.a); 54 // dot1=cross(l1.b-l2.a , l1.a-l2.a); 55 // //dot2 = multi(l1.b, l2.b, l1.a); 56 // dot2=cross(l2.b-l1.b , l1.a-l1.b); 57 // p.x = (l2.a.x * dot2 + l2.b.x * dot1) / (dot2 + dot1); 58 // p.y = (l2.a.y * dot2 + l2.b.y * dot1) / (dot2 + dot1); 59 // return p; 60 //} 61 int HalfplaneIntersection(Line *L,int n,Point *poly) 62 { 63 sort(L,L+n); 64 65 int first,last; 66 Point *p=new Point[n]; 67 Line *q=new Line[n]; 68 q[first=last=0]=L[0]; 69 for (int i=1 ;i<n ;i++) 70 { 71 while (first<last && !OnLeft(L[i],p[last-1])) last--; 72 while (first<last && !OnLeft(L[i],p[first])) first++; 73 q[++last]=L[i]; 74 if (fabs(cross(q[last].v , q[last-1].v))<exp) 75 { 76 last--; 77 if (OnLeft(q[last] , L[i].p)) q[last]=L[i]; 78 } 79 if (first<last) p[last-1]=GetIntersection(q[last-1],q[last]); 80 } 81 while (first<last && !OnLeft(q[first],p[last-1])) last--; 82 if (last-first<=1) return 0; 83 p[last]=GetIntersection(q[last],q[first]); 84 int m=0; 85 for (int i=first ;i<=last ;i++) poly[m++]=p[i]; 86 return m; 87 } 88 void calPolygon(Point *p, int n, double &area, bool &shun) 89 { 90 p[n] = p[0]; 91 area = 0; 92 double tmp; 93 for (int i = 0; i < n; i++) 94 area += p[i].x * p[i + 1].y - p[i].y * p[i + 1].x; 95 area /= 2.0; 96 if (shun = area < 0) 97 area = -area; 98 } 99 bool calCore(Point *ps, int n) 100 { 101 Line l[maxn]; 102 ps[n] = ps[0]; 103 bool shun; 104 double area; 105 calPolygon(ps, n, area, shun); 106 if (shun) 107 for (int i = 0; i < n; i++) 108 bn[i] = Line(ps[i], ps[i] - ps[i + 1]); 109 else 110 for (int i = 0; i < n; i++) 111 bn[i] = Line(ps[i], ps[i + 1] - ps[i]); 112 Point pp[maxn]; 113 return HalfplaneIntersection(bn, n, pp); 114 } 115 int main() 116 { 117 int t; 118 int n; 119 scanf("%d",&t); 120 while (t--) 121 { 122 scanf("%d",&n); 123 Point cn[maxn]; 124 for (int i=0 ;i<n ;i++) 125 { 126 scanf("%lf%lf",&cn[i].x,&cn[i].y); 127 } 128 // reverse(cn,cn+n); 129 // for (int i=0 ;i<n ;i++) 130 // { 131 // bn[i].p=cn[i]; 132 // bn[i].v=cn[(i+1)%n]-cn[i]; 133 // bn[i].ang=atan2(bn[i].v.y , bn[i].v.x); 134 // } 135 // int m=HalfplaneIntersection(bn,n,an); 136 if (!calCore(cn,n)) puts("NO"); 137 else puts("YES"); 138 } 139 return 0; 140 }
练习：
后续：欢迎提出宝贵的意见。
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