题目大意:
有许多三角形可能互相重叠
分别输出重叠了n片到只重叠了1片的面积
解法1:官方题解切多边形,木有尝试
解法2:即以下代码,参考http://www.cnblogs.com/woaishizhan/archive/2013/08/01/3229234.html
解法3:本人尝试用PSLG生成所有互不覆盖的多边形
然后遍历每个三角形判断多边形是否在三角形内
最后累计所有多边形的面积 最终没有调出来
有人有PSLG生成多边形的模板否。。。。。感激不尽
//大白p263
#include <cmath>
#include <cstdio>
#include <cstring>
#include <set>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
const double eps=1e-9;//精度
const int INF=1<<29;
const double PI=acos(-1.0);
int dcmp(double x){//判断double等于0或。。。
if(fabs(x)<eps)return 0;else return x<0?-1:1;
}
struct Point{
double x,y;
Point(double x=0,double y=0):x(x),y(y){}
};
typedef Point Vector;
Vector operator+(Vector a,Vector b){return Vector(a.x+b.x,a.y+b.y);}//向量+向量=向量
Vector operator-(Point a,Point b){return Vector(a.x-b.x,a.y-b.y);}//点-点=向量
Vector operator*(Vector a,double p){return Vector(a.x*p,a.y*p);}//向量*实数=向量
Vector operator/(Vector a,double p){return Vector(a.x/p,a.y/p);}//向量/实数=向量
bool operator<( const Point& A,const Point& B ){return dcmp(A.x-B.x)<0||(dcmp(A.x-B.x)==0&&dcmp(A.y-B.y)<0);}
bool operator==(const Point&a,const Point&b){return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;}
bool operator!=(const Point&a,const Point&b){return a==b;}
struct Segment{
Point a,b;
Segment(){}
Segment(Point _a,Point _b){a=_a,b=_b;}
bool friend operator<(const Segment& p,const Segment& q){return p.a<q.a||(p.a==q.a&&p.b<q.b);}
bool friend operator==(const Segment& p,const Segment& q){return (p.a==q.a&&p.b==q.b)||(p.a==q.b&&p.b==q.a);}
};
struct Circle{
Point c;
double r;
Circle(){}
Circle(Point _c, double _r):c(_c),r(_r) {}
Point point(double a)const{return Point(c.x+cos(a)*r,c.y+sin(a)*r);}
bool friend operator<(const Circle& a,const Circle& b){return a.r<b.r;}
};
struct Line{
Point p;
Vector v;
double ang;
Line() {}
Line(const Point &_p, const Vector &_v):p(_p),v(_v){ang = atan2(v.y, v.x);}
bool operator<(const Line &L)const{return ang < L.ang;}
};
double Dot(Vector a,Vector b){return a.x*b.x+a.y*b.y;}//|a|*|b|*cosθ 点积
double Length(Vector a){return sqrt(Dot(a,a));}//|a| 向量长度
double Angle(Vector a,Vector b){return acos(Dot(a,b)/Length(a)/Length(b));}//向量夹角θ
double Cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;}//叉积 向量围成的平行四边形的面积
double Area2(Point a,Point b,Point c){return Cross(b-a,c-a);}//同上 参数为三个点
double DegreeToRadius(double deg){return deg/180*PI;}
double GetRerotateAngle(Vector a,Vector b){//向量a顺时针旋转theta度得到向量b的方向
double tempa=Angle(a,Vector(1,0));
if(a.y<0) tempa=2*PI-tempa;
double tempb=Angle(b,Vector(1,0));
if(b.y<0) tempb=2*PI-tempb;
if((tempa-tempb)>0) return tempa-tempb;
else return tempa-tempb+2*PI;
}
Vector Rotate(Vector a,double rad){//向量旋转rad弧度
return Vector(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
}
Vector Normal(Vector a){//计算单位法线
double L=Length(a);
return Vector(-a.y/L,a.x/L);
}
Point GetLineProjection(Point p,Point a,Point b){//点在直线上的投影
Vector v=b-a;
return a+v*(Dot(v,p-a)/Dot(v,v));
}
Point GetLineIntersection(Point p,Vector v,Point q,Vector w){//求直线交点 有唯一交点时可用
Vector u=p-q;
double t=Cross(w,u)/Cross(v,w);
return p+v*t;
}
int ConvexHull(Point* p,int n,Point* ch){//计算凸包
sort(p,p+n);
int m=0;
for(int i=0;i<n;i++){
while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
ch[m++]=p[i];
}
int k=m;
for(int i=n-2;i>=0;i--){
while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
ch[m++]=p[i];
}
if(n>0) m--;
return m;
}
double Heron(double a,double b,double c){//海伦公式
double p=(a+b+c)/2;
return sqrt(p*(p-a)*(p-b)*(p-c));
}
bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2){//线段规范相交判定
double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1);
double c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);
return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
}
double CutConvex(const int n,Point* poly, const Point a,const Point b, vector<Point> result[3]){//有向直线a b 切割凸多边形
vector<Point> points;
Point p;
Point p1=a,p2=b;
int cur,pre;
result[0].clear();
result[1].clear();
result[2].clear();
if(n==0) return 0;
double tempcross;
tempcross=Cross(p2-p1,poly[0]-p1);
if(dcmp(tempcross)==0) pre=cur=2;
else if(tempcross>0) pre=cur=0;
else pre=cur=1;
for(int i=0;i<n;i++){
tempcross=Cross(p2-p1,poly[(i+1)%n]-p1);
if(dcmp(tempcross)==0) cur=2;
else if(tempcross>0) cur=0;
else cur=1;
if(cur==pre){
result[cur].push_back(poly[(i+1)%n]);
}
else{
p1=poly[i];
p2=poly[(i+1)%n];
p=GetLineIntersection(p1,p2-p1,a,b-a);
points.push_back(p);
result[pre].push_back(p);
result[cur].push_back(p);
result[cur].push_back(poly[(i+1)%n]);
pre=cur;
}
}
sort(points.begin(),points.end());
if(points.size()<2){
return 0;
}
else{
return Length(points.front()-points.back());
}
}
double DistanceToSegment(Point p,Segment s){//点到线段的距离
if(s.a==s.b) return Length(p-s.a);
Vector v1=s.b-s.a,v2=p-s.a,v3=p-s.b;
if(dcmp(Dot(v1,v2))<0) return Length(v2);
else if(dcmp(Dot(v1,v3))>0) return Length(v3);
else return fabs(Cross(v1,v2))/Length(v1);
}
bool isPointOnSegment(Point p,Segment s){//点在线段上
return dcmp(DistanceToSegment(p,s))==0;
}
int isPointInPolygon(Point p, Point* poly,int n){//点与多边形的位置关系
int wn=0;
for(int i=0;i<n;i++){
Point& p2=poly[(i+1)%n];
if(isPointOnSegment(p,Segment(poly[i],p2))) return -1;//点在边界上
int k=dcmp(Cross(p2-poly[i],p-poly[i]));
int d1=dcmp(poly[i].y-p.y);
int d2=dcmp(p2.y-p.y);
if(k>0&&d1<=0&&d2>0)wn++;
if(k<0&&d2<=0&&d1>0)wn--;
}
if(wn) return 1;//点在内部
else return 0;//点在外部
}
double PolygonArea(vector<Point> p){//多边形有向面积
double area=0;
int n=p.size();
for(int i=1;i<n-1;i++)
area+=Cross(p[i]-p[0],p[i+1]-p[0]);
return area/2;
}
int GetLineCircleIntersection(Line L,Circle C,Point& p1,Point& p2){//圆与直线交点 返回交点个数
double a = L.v.x, b = L.p.x - C.c.x, c = L.v.y, d = L.p.y-C.c.y;
double e = a*a + c*c, f = 2*(a*b+c*d), g = b*b + d*d -C.r*C.r;
double delta = f*f - 4*e*g;
if(dcmp(delta) < 0) return 0;//相离
if(dcmp(delta) == 0) {//相切
p1=p1=C.point(-f/(2*e));
return 1;
}//相交
p1=(L.p+L.v*(-f-sqrt(delta))/(2*e));
p2=(L.p+L.v*(-f+sqrt(delta))/(2*e));
return 2;
}
//--------------------------------------
//--------------------------------------
//--------------------------------------
//--------------------------------------
//--------------------------------------
Point parr[9999];
Segment sarr[1600];
bool smark[1600];
int pidx,sidx;
struct Segforcmp{
Segment s;
bool mark;
bool friend operator<(Segforcmp a,Segforcmp b){
return a.s<b.s;
}
}temps[2000];
bool isintri(Point m,Point a,Point b){
Point p=GetLineProjection(m,a,b);
if(m.y>p.y) return true;
else return false;
}
int main()
{
int T,n;
scanf("%d",&T);
while(T--){
sidx=pidx=0;
scanf("%d",&n);
double ans[100];
memset(ans,0,sizeof ans);
//输入处理
for(int i=0;i<n;i++){
Point a,b,c,m;
scanf("%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y);
m.x=(a.x+b.x+c.x)/3;
m.y=(a.y+b.y+c.y)/3;
if(dcmp(Cross(a-b,b-c))==0) continue;//共线舍去
parr[pidx++]=a;parr[pidx++]=b;parr[pidx++]=c;//端点放入数组
if(a.x<b.x) sarr[sidx]=Segment(a,b);
else sarr[sidx]=Segment(b,a);
smark[sidx++]=isintri(m,a,b);//边放入数组 并标记出入边
if(a.x<c.x) sarr[sidx]=Segment(a,c);
else sarr[sidx]=Segment(c,a);
smark[sidx++]=isintri(m,a,c);
if(b.x<c.x) sarr[sidx]=Segment(b,c);
else sarr[sidx]=Segment(c,b);
smark[sidx++]=isintri(m,c,b);
}
//两两边求交点
for(int i=0;i<sidx;i++){
for(int j=i+1;j<sidx;j++){
if(SegmentProperIntersection(sarr[i].a,sarr[i].b,sarr[j].a,sarr[j].b))
parr[pidx++]=GetLineIntersection(sarr[i].a,sarr[i].b-sarr[i].a,sarr[j].a,sarr[j].b-sarr[j].a);
}
}
sort(parr,parr+pidx);
Point pre=parr[0],cur;
for(int i=1;i<pidx;i++)if(dcmp(parr[i].x-pre.x)!=0){
cur=parr[i];
int tempscount=0;
for(int j=0;j<sidx;j++)if(sarr[j].a.x<=pre.x&&sarr[j].b.x>=cur.x){
Point a=GetLineIntersection(pre,Vector(0,1),sarr[j].a,sarr[j].b-sarr[j].a);//左边扫描线与线段交点
Point b=GetLineIntersection(cur,Vector(0,1),sarr[j].a,sarr[j].b-sarr[j].a);//右边。。。。
Segment temp(a,b);
temps[tempscount].s=temp;
temps[tempscount++].mark=smark[j];
}
sort(temps,temps+tempscount);
int deep=temps[0].mark;
for(int j=1;j<tempscount;j++){//遍历区间中的线段求面积
double h=temps[j].s.b.x-temps[j].s.a.x;
double base=fabs(temps[j-1].s.a.y-temps[j].s.a.y)+fabs(temps[j-1].s.b.y-temps[j].s.b.y);
if(deep) ans[deep]+=base/2*h;
if(temps[j].mark) deep++;
else deep--;
}
pre=cur;
}
for(int i=1;i<=n;i++)
printf("%.10lf\n",ans[i]);
}
return 0;
}