题意： 一个(0,0)到(10,10)的矩形，目标点不定，从(0,0)开始走，如果走到新一点是"Hotter",那么意思是离目标点近了，如果是"Colder“，那么就是远了，"Same"是相同。要你推测目标点的可能位置的面积。
解法：半平面交水题。从一个点到另一个点远了，说明目标点在两点之间连线的中垂线的离源点较近的一侧，即我们每次都可以得到一条直线来切割平面，要么切割左侧，要么切割右侧，要么都切，再求一个半平面交就可以得出可能面积了。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define pi acos(-1.0)
#define eps 1e-8
using namespace std; struct Point{
double x,y;
Point(double x=, double y=):x(x),y(y) {}
void input() { scanf("%lf%lf",&x,&y); }
};
typedef Point Vector;
struct Line{
Point p;
Vector v;
double ang;
Line(){}
Line(Point p, Vector v):p(p),v(v) { ang = atan2(v.y,v.x); }
Point point(double t) { return Point(p.x + t*v.x, p.y + t*v.y); }
bool operator < (const Line &L)const { return ang < L.ang; }
};
int dcmp(double x) {
if(x < -eps) return -;
if(x > eps) return ;
return ;
}
template <class T> T sqr(T x) { return x * x;}
Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }
bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; }
bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; }
bool operator == (const Point& a, const Point& b) { return dcmp(a.x-b.x) == && dcmp(a.y-b.y) == ; }
double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
Vector VectorUnit(Vector x){ return x / Length(x);}
Vector Normal(Vector x) { return Point(-x.y, x.x) / Length(x);}
double angle(Vector v) { return atan2(v.y, v.x); } Point GetLineIntersection(Line A, Line B) {
Vector u = A.p - B.p;
double t = Cross(B.v, u) / Cross(A.v, B.v);
return A.p + A.v*t;
}
double DisP(Point A,Point B) {
return Length(B-A);
}
double CalcConvexArea(Point* p,int n) { //凸包面积
double area = 0.0;
for(int i=;i<n-;i++)
area += Cross(p[i]-p[],p[i+]-p[]);
return fabs(area*0.5);
}
bool OnLeft(Line L, Point p) { return dcmp(Cross(L.v,p-L.p)) > ; }
bool CmpPolarLine(Line a,Line b) { //直线极角排序
return angle(a.v) < angle(b.v);
}
int HalfPlaneIntersection(Line* L, int n, Point* poly) { //半平面交点存入poly
sort(L,L+n,CmpPolarLine);
int first,last;
Point *p = new Point[n];
Line *q = new Line[n];
q[first=last=] = L[];
for(int i=;i<n;i++) {
while(first < last && !OnLeft(L[i],p[last-])) last--;
while(first < last && !OnLeft(L[i],p[first])) first++;
q[++last] = L[i];
if(dcmp(Cross(q[last].v, q[last-].v)) == ) {
last--;
if(OnLeft(q[last], L[i].p)) q[last] = L[i];
}
if(first < last) p[last-] = GetLineIntersection(q[last-],q[last]);
}
while(first < last && !OnLeft(q[first],p[last-])) last--;
if(last-first <= ) return ; //点或线或*平面，返回0
p[last] = GetLineIntersection(q[last],q[first]);
int m = ;
for(int i=first;i<=last;i++) poly[m++] = p[i];
delete p; delete q;
return m;
} Line L[],TL[];
Point poly[]; int main()
{
int i,j,tot = -;
Point n,p;
char ss[];
p.x = p.y = 0.0;
TL[++tot] = Line(Point(,),Vector(,));
TL[++tot] = Line(Point(,),Vector(,));
TL[++tot] = Line(Point(,),Vector(-,));
TL[++tot] = Line(Point(,),Vector(,-));
while(scanf("%lf%lf%s",&n.x,&n.y,ss)!=EOF)
{
if(ss[] == 'H')
TL[++tot] = Line(Point((n.x+p.x)/2.0,(n.y+p.y)/2.0),Vector(Normal(p-n)));
else if(ss[] == 'C')
TL[++tot] = Line(Point((n.x+p.x)/2.0,(n.y+p.y)/2.0),Vector(Normal(n-p)));
else {
TL[++tot] = Line(Point((n.x+p.x)/2.0,(n.y+p.y)/2.0),Vector(Normal(p-n)));
TL[++tot] = Line(Point((n.x+p.x)/2.0,(n.y+p.y)/2.0),Vector(Normal(n-p)));
}
p = n;
for(i=;i<=tot;i++) L[i] = TL[i];
int m = HalfPlaneIntersection(L,tot+,poly);
if(!m) puts("0.00");
else printf("%.2f\n",CalcConvexArea(poly,m));
}
return ;
}