半平面交算法及简单应用

                              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 }

View Code

 

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 }

View Code

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 }

View Code

 

 

练习：

poj 1474 

poj 2451

poj 3525

LA 2218

LA 2512

UVA 10084

 

 

后续：欢迎提出宝贵的意见。