半平面交

5.1  半平面交模板(from UESTC) 
 
const double eps = 1e-8; 
const double PI = acos(-1.0); 
int sgn(double x) 
{ 
  if(fabs(x) < eps) return 0; 
  if(x < 0) return -1; 
  else return 1; 
} 
struct Point 
{ 
  double x,y; 
  Point(){} 
  Point(double _x,double _y) 
  { 
    x = _x; y = _y; 
  } 
  Point operator -(const Point &b)const 
  { 
    return Point(x - b.x, y - b.y); 
  } 
  double operator ^(const Point &b)const 
  { 
    return x*b.y - y*b.x; 
  } 
  double operator *(const Point &b)const 
  { 
    return x*b.x + y*b.y; 
  } 
}; 
struct Line 
{ 
  Point s,e; 
  double k; 
  Line(){} 
  Line(Point _s,Point _e) 
  { 
    s = _s; e = _e; 
    k = atan2(e.y - s.y,e.x - s.x); 
  } 
  Point operator &(const Line &b)const 
  { 
    Point res = s; 
    double t = ((s - b.s)^(b.s - b.e))/((s - e)^(b.s - b.e)); 
    res.x += (e.x - s.x)*t; 
    res.y += (e.y - s.y)*t; 
    return res; 
  } 
}; 
//半平面交，直线的左边代表有效区域 
bool HPIcmp(Line a,Line b) 
{ 
  if(fabs(a.k - b.k) > eps)return a.k < b.k; 
  return ((a.s - b.s)^(b.e - b.s)) < 0; 
} 
Line Q[110]; 
void HPI(Line line[], int n, Point res[], int &resn) 
{ 
  int tot = n; 
  sort(line,line+n,HPIcmp); 
  tot = 1; 
  for(int i = 1;i < n;i++) 
    if(fabs(line[i].k - line[i-1].k) > eps)
      line[tot++] = line[i]; 
  int head = 0, tail = 1; 
  Q[0] = line[0]; 
  Q[1] = line[1]; 
  resn = 0; 
  for(int i = 2; i < tot; i++) 
  { 
    if(fabs((Q[tail].e-Q[tail].s)^(Q[tail-1].e-Q[tail-1].s)) < eps || 
fabs((Q[head].e-Q[head].s)^(Q[head+1].e-Q[head+1].s)) < eps) 
      return; 
    while(head < tail && (((Q[tail]&Q[tail-1]) - 
line[i].s)^(line[i].e-line[i].s)) > eps) 
      tail--; 
    while(head < tail && (((Q[head]&Q[head+1]) - 
line[i].s)^(line[i].e-line[i].s)) > eps) 
      head++; 
    Q[++tail] = line[i]; 
  } 
  while(head < tail && (((Q[tail]&Q[tail-1]) - 
Q[head].s)^(Q[head].e-Q[head].s)) > eps) 
    tail--; 
  while(head < tail && (((Q[head]&Q[head-1]) - 
Q[tail].s)^(Q[tail].e-Q[tail].e)) > eps) 
    head++; 
  if(tail <= head + 1)return; 
  for(int i = head; i < tail; i++) 
    res[resn++] = Q[i]&Q[i+1]; 
  if(head < tail - 1) 
    res[resn++] = Q[head]&Q[tail]; 
} 
5.2  普通半平面交写法 
POJ 1750 
const double eps = 1e-18; 
int sgn(double x) 
{ 
  if(fabs(x) < eps)return 0; 
  if(x < 0)return -1; 
  else return 1; 
} 
struct Point 
{ 
  double x,y; 
  Point(){} 
  Point(double _x,double _y) 
  { 
    x = _x; y = _y; 
  } 
  Point operator -(const Point &b)const 
  { 
    return Point(x - b.x, y - b.y); 
  } 
  double operator ^(const Point &b)const 
  { 
    return x*b.y - y*b.x; 
  } 
  double operator *(const Point &b)const 
  { 
    return x*b.x + y*b.y; 
  } 
}; 
//计算多边形面积 
double CalcArea(Point p[],int n) 
{ 
  double res = 0; 
  for(int i = 0;i < n;i++) 
    res += (p[i]^p[(i+1)%n]); 
  return fabs(res/2); 
} 
//通过两点，确定直线方程 
void Get_equation(Point p1,Point p2,double &a,double &b,double &c) 
{ 
  a = p2.y - p1.y; 
  b = p1.x - p2.x; 
  c = p2.x*p1.y - p1.x*p2.y; 
} 
//求交点 
Point Intersection(Point p1,Point p2,double a,double b,double c) 
{ 
  double u = fabs(a*p1.x + b*p1.y + c); 
  double v = fabs(a*p2.x + b*p2.y + c); 
  Point t; 
  t.x = (p1.x*v + p2.x*u)/(u+v); 
  t.y = (p1.y*v + p2.y*u)/(u+v); 
  return t; 
} 
Point tp[110]; 
void Cut(double a,double b,double c,Point p[],int &cnt) 
{ 
  int tmp = 0; 
  for(int i = 1;i <= cnt;i++) 
  { 
    //当前点在左侧，逆时针的点 
    if(a*p[i].x + b*p[i].y + c < eps)tp[++tmp] = p[i]; 
    else 
    { 
      if(a*p[i-1].x + b*p[i-1].y + c < -eps) 
        tp[++tmp] = Intersection(p[i-1],p[i],a,b,c); 
      if(a*p[i+1].x + b*p[i+1].y + c < -eps) 
        tp[++tmp] = Intersection(p[i],p[i+1],a,b,c); 
    } 
  } 
  for(int i = 1;i <= tmp;i++) 
    p[i] = tp[i]; 
  p[0] = p[tmp]; 
  p[tmp+1] = p[1]; 
  cnt = tmp; 
} 
double V[110],U[110],W[110]; 
int n; 
const double INF = 100000000000.0; 
Point p[110]; 
bool solve(int id) 
{ 
  p[1] = Point(0,0); 
  p[2] = Point(INF,0); 
  p[3] = Point(INF,INF); 
  p[4] = Point(0,INF); 
  p[0] = p[4]; 
  p[5] = p[1]; 
  int cnt = 4; 
  for(int i = 0;i < n;i++) 
    if(i != id) 
    { 
      double a = (V[i] - V[id])/(V[i]*V[id]); 
      double b = (U[i] - U[id])/(U[i]*U[id]); 
      double c = (W[i] - W[id])/(W[i]*W[id]); 
      if(sgn(a) == 0 && sgn(b) == 0) 
      { 
        if(sgn(c) >= 0)return false; 
        else continue; 
      } 
      Cut(a,b,c,p,cnt); 
    } 
  if(sgn(CalcArea(p,cnt)) == 0)return false; 
  else return true; 
} 
int main() 
{ 
    while(scanf("%d",&n) == 1) 
  { 
    for(int i = 0;i < n;i++) 
      scanf("%lf%lf%lf",&V[i],&U[i],&W[i]); 
    for(int i = 0;i < n;i++) 
    { 
      if(solve(i))printf("Yes
"); 
      else printf("No
"); 
    } 
  } 
    return 0; 
}