Problem Description
This is a simple problem. Given two triangles A and B, you should determine they are intersect, contain or disjoint. (Public edge or point are treated as intersect.)
Input
First line contains an integer T (1 ≤ T ≤ 10), represents there are T test cases.
For each test case: X1 Y1 X2 Y2 X3 Y3 X4 Y4 X5 Y5 X6 Y6. All the coordinate are integer. (X1,Y1) , (X2,Y2), (X3,Y3) forms triangles A ; (X4,Y4) , (X5,Y5), (X6,Y6) forms triangles B.
-10000<=All the coordinate <=10000
Output
For each test case, output “intersect”, “contain” or “disjoint”.
Sample Input
2
0 0 0 1 1 0 10 10 9 9 9 10
0 0 1 1 1 0 0 0 1 1 0 1
Sample Output
disjoint
intersect
观察样例可以发现，如果边有重叠部分，此题也算相交。
因此我套用多边形面积交的模板http://www.cnblogs.com/Annetree/p/6535294.html
如果有重合面积，有两种情况：包含或相交，特殊判断包含即可
如果没有重合面积，也有两种情况：相交（线）和相离，特殊判断线部分重合即可
#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<queue>
#include<map>
#include<stack>
#include<set> using namespace std; const int maxn=;
const int maxisn=;
const double eps=1e-;
const double pi=acos(-1.0); int dcmp(double x){
if(x>eps) return ;
return x<-eps ? - : ;
}
inline double Sqr(double x){
return x*x;
} #define zero(x)(((x)>0?(x):-(x))<eps)
struct Point{
double x,y;
Point(){x=y=;}
Point(double x,double y):x(x),y(y){};
friend Point operator + (const Point &a,const Point &b) {
return Point(a.x+b.x,a.y+b.y);
}
friend Point operator - (const Point &a,const Point &b) {
return Point(a.x-b.x,a.y-b.y);
}
friend bool operator == (const Point &a,const Point &b) {
return dcmp(a.x-b.x)==&&dcmp(a.y-b.y)==;
}
friend Point operator * (const Point &a,const double &b) {
return Point(a.x*b,a.y*b);
}
friend Point operator * (const double &a,const Point &b) {
return Point(a*b.x,a*b.y);
}
friend Point operator / (const Point &a,const double &b) {
return Point(a.x/b,a.y/b);
}
friend bool operator < (const Point &a, const Point &b) {
return a.x < b.x || (a.x == b.x && a.y < b.y);
}
inline double dot(const Point &b)const{
return x*b.x+y*b.y;
}
inline double cross(const Point &b,const Point &c)const{
return (b.x-x)*(c.y-y)-(c.x-x)*(b.y-y);
} }; Point LineCross(const Point &a,const Point &b,const Point &c,const Point &d){
double u=a.cross(b,c),v=b.cross(a,d);
return Point((c.x*v+d.x*u)/(u+v),(c.y*v+d.y*u)/(u+v));
}
double PolygonArea(Point p[],int n){
if(n<) return 0.0;
double s=p[].y*(p[n-].x-p[].x);
p[n]=p[];
for(int i=;i<n;i++){
s+=p[i].y*(p[i-].x-p[i+].x);
}
return fabs(s*0.5);
}
double CPIA(Point a[],Point b[],int na,int nb){
Point p[maxisn],temp[maxisn];
int i,j,tn,sflag,eflag;
a[na]=a[],b[nb]=b[];
memcpy(p,b,sizeof(Point)*(nb+));
for(i=;i<na&&nb>;++i){
sflag=dcmp(a[i].cross(a[i+],p[]));
for(j=tn=;j<nb;++j,sflag=eflag){
if(sflag>=) temp[tn++]=p[j];
eflag=dcmp(a[i].cross(a[i+],p[j+]));
if((sflag^eflag)==-)
temp[tn++]=LineCross(a[i],a[i+],p[j],p[j+]);
}
memcpy(p,temp,sizeof(Point)*tn);
nb=tn,p[nb]=p[];
}
if(nb<) return 0.0;
return PolygonArea(p,nb);
}
double SPIA(Point a[],Point b[],int na,int nb){
int i,j;
Point t1[],t2[];
double res=0.0,if_clock_t1,if_clock_t2;
a[na]=t1[]=a[];
b[nb]=t2[]=b[];
for(i=;i<na;i++){
t1[]=a[i-],t1[]=a[i];
if_clock_t1=dcmp(t1[].cross(t1[],t1[]));
if(if_clock_t1<) swap(t1[],t1[]);
for(j=;j<nb;j++){
t2[]=b[j-],t2[]=b[j];
if_clock_t2=dcmp(t2[].cross(t2[],t2[]));
if(if_clock_t2<) swap(t2[],t2[]);
res+=CPIA(t1,t2,,)*if_clock_t1*if_clock_t2;
}
}
return res;
//return PolygonArea(a,na)+PolygonArea(b,nb)-res;
} Point a[],b[];
Point aa[],bb[]; double length(Point p1,Point p2)//求边长
{
double res=sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
return res;
} double area_triangle(double l1,double l2,double l3)//求三角形面积
{
double s=(l1+l2+l3)/2.0;
double res=sqrt(s*(s-l1)*(s-l2)*(s-l3));
return res;
} double xmult(Point p1,Point p2,Point p0)
{
return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
} int parallel(Point u1,Point u2,Point v1,Point v2)//判断平行
{
return zero((u1.x-u2.x)*(v1.y-v2.y)-(v1.x-v2.x)*(u1.y-u2.y));
} int dot_online_in(Point p,Point l1,Point l2)
{
return zero(xmult(p,l1,l2))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
for(int i=;i<;i++) scanf("%lf %lf",&a[i].x,&a[i].y);
for(int i=;i<;i++) scanf("%lf %lf",&b[i].x,&b[i].y);
double area_double =fabs(SPIA(a,b,,));//重合面积
double l1=length(a[],a[]),l2=length(a[],a[]),l3=length(a[],a[]);
double l4=length(b[],b[]),l5=length(b[],b[]),l6=length(b[],b[]);
if(area_double>eps) //包含或相交
{
if(abs(area_double-area_triangle(l1,l2,l3))<eps)
printf("contain\n");
else if(abs(area_double-area_triangle(l4,l5,l6))<eps)
printf("contain\n");
else
printf("intersect\n");
}
else //相交或相离
{
bool flag=false;
//判断是否有边重合
for(int i=;i<;i++)
{
for(int ii=i+;ii<;ii++)
{
for(int j=;j<;j++)
{
for(int jj=j+;jj<;jj++)
{
if(parallel(a[i],a[ii],b[j],b[jj]))
if(dot_online_in(a[i],b[j],b[jj]))
{
flag=true;
break;
}
}
if(flag)
break;
}
if(flag)
break;
}
if(flag)
break;
}
if(flag)
printf("intersect\n");
else
printf("disjoint\n"); }
}
return ;
}