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
1 #include<cstdio> 2 #include<cmath> 3 #include<cstring> 4 #include<iostream> 5 #include<algorithm> 6 #include<cstdlib> 7 #include<queue> 8 #include<map> 9 #include<stack> 10 #include<set> 11 12 using namespace std; 13 14 const int maxn=555; 15 const int maxisn=10; 16 const double eps=1e-8; 17 const double pi=acos(-1.0); 18 19 int dcmp(double x){ 20 if(x>eps) return 1; 21 return x<-eps ? -1 : 0; 22 } 23 inline double Sqr(double x){ 24 return x*x; 25 } 26 27 #define zero(x)(((x)>0?(x):-(x))<eps) 28 struct Point{ 29 double x,y; 30 Point(){x=y=0;} 31 Point(double x,double y):x(x),y(y){}; 32 friend Point operator + (const Point &a,const Point &b) { 33 return Point(a.x+b.x,a.y+b.y); 34 } 35 friend Point operator - (const Point &a,const Point &b) { 36 return Point(a.x-b.x,a.y-b.y); 37 } 38 friend bool operator == (const Point &a,const Point &b) { 39 return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0; 40 } 41 friend Point operator * (const Point &a,const double &b) { 42 return Point(a.x*b,a.y*b); 43 } 44 friend Point operator * (const double &a,const Point &b) { 45 return Point(a*b.x,a*b.y); 46 } 47 friend Point operator / (const Point &a,const double &b) { 48 return Point(a.x/b,a.y/b); 49 } 50 friend bool operator < (const Point &a, const Point &b) { 51 return a.x < b.x || (a.x == b.x && a.y < b.y); 52 } 53 inline double dot(const Point &b)const{ 54 return x*b.x+y*b.y; 55 } 56 inline double cross(const Point &b,const Point &c)const{ 57 return (b.x-x)*(c.y-y)-(c.x-x)*(b.y-y); 58 } 59 60 }; 61 62 Point LineCross(const Point &a,const Point &b,const Point &c,const Point &d){ 63 double u=a.cross(b,c),v=b.cross(a,d); 64 return Point((c.x*v+d.x*u)/(u+v),(c.y*v+d.y*u)/(u+v)); 65 } 66 double PolygonArea(Point p[],int n){ 67 if(n<3) return 0.0; 68 double s=p[0].y*(p[n-1].x-p[1].x); 69 p[n]=p[0]; 70 for(int i=1;i<n;i++){ 71 s+=p[i].y*(p[i-1].x-p[i+1].x); 72 } 73 return fabs(s*0.5); 74 } 75 double CPIA(Point a[],Point b[],int na,int nb){ 76 Point p[maxisn],temp[maxisn]; 77 int i,j,tn,sflag,eflag; 78 a[na]=a[0],b[nb]=b[0]; 79 memcpy(p,b,sizeof(Point)*(nb+1)); 80 for(i=0;i<na&&nb>2;++i){ 81 sflag=dcmp(a[i].cross(a[i+1],p[0])); 82 for(j=tn=0;j<nb;++j,sflag=eflag){ 83 if(sflag>=0) temp[tn++]=p[j]; 84 eflag=dcmp(a[i].cross(a[i+1],p[j+1])); 85 if((sflag^eflag)==-2) 86 temp[tn++]=LineCross(a[i],a[i+1],p[j],p[j+1]); 87 } 88 memcpy(p,temp,sizeof(Point)*tn); 89 nb=tn,p[nb]=p[0]; 90 } 91 if(nb<3) return 0.0; 92 return PolygonArea(p,nb); 93 } 94 double SPIA(Point a[],Point b[],int na,int nb){ 95 int i,j; 96 Point t1[4],t2[4]; 97 double res=0.0,if_clock_t1,if_clock_t2; 98 a[na]=t1[0]=a[0]; 99 b[nb]=t2[0]=b[0]; 100 for(i=2;i<na;i++){ 101 t1[1]=a[i-1],t1[2]=a[i]; 102 if_clock_t1=dcmp(t1[0].cross(t1[1],t1[2])); 103 if(if_clock_t1<0) swap(t1[1],t1[2]); 104 for(j=2;j<nb;j++){ 105 t2[1]=b[j-1],t2[2]=b[j]; 106 if_clock_t2=dcmp(t2[0].cross(t2[1],t2[2])); 107 if(if_clock_t2<0) swap(t2[1],t2[2]); 108 res+=CPIA(t1,t2,3,3)*if_clock_t1*if_clock_t2; 109 } 110 } 111 return res; 112 //return PolygonArea(a,na)+PolygonArea(b,nb)-res; 113 } 114 115 Point a[222],b[222]; 116 Point aa[222],bb[222]; 117 118 double length(Point p1,Point p2)//求边长 119 { 120 double res=sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)); 121 return res; 122 } 123 124 double area_triangle(double l1,double l2,double l3)//求三角形面积 125 { 126 double s=(l1+l2+l3)/2.0; 127 double res=sqrt(s*(s-l1)*(s-l2)*(s-l3)); 128 return res; 129 } 130 131 double xmult(Point p1,Point p2,Point p0) 132 { 133 return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); 134 } 135 136 int parallel(Point u1,Point u2,Point v1,Point v2)//判断平行 137 { 138 return zero((u1.x-u2.x)*(v1.y-v2.y)-(v1.x-v2.x)*(u1.y-u2.y)); 139 } 140 141 int dot_online_in(Point p,Point l1,Point l2) 142 { 143 return zero(xmult(p,l1,l2))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps; 144 } 145 int main() 146 { 147 int T; 148 scanf("%d",&T); 149 while(T--) 150 { 151 for(int i=0;i<3;i++) scanf("%lf %lf",&a[i].x,&a[i].y); 152 for(int i=0;i<3;i++) scanf("%lf %lf",&b[i].x,&b[i].y); 153 double area_double =fabs(SPIA(a,b,3,3));//重合面积 154 double l1=length(a[0],a[1]),l2=length(a[0],a[2]),l3=length(a[1],a[2]); 155 double l4=length(b[0],b[1]),l5=length(b[0],b[2]),l6=length(b[1],b[2]); 156 if(area_double>eps) //包含或相交 157 { 158 if(abs(area_double-area_triangle(l1,l2,l3))<eps) 159 printf("contain "); 160 else if(abs(area_double-area_triangle(l4,l5,l6))<eps) 161 printf("contain "); 162 else 163 printf("intersect "); 164 } 165 else //相交或相离 166 { 167 bool flag=false; 168 //判断是否有边重合 169 for(int i=0;i<2;i++) 170 { 171 for(int ii=i+1;ii<3;ii++) 172 { 173 for(int j=0;j<2;j++) 174 { 175 for(int jj=j+1;jj<3;jj++) 176 { 177 if(parallel(a[i],a[ii],b[j],b[jj])) 178 if(dot_online_in(a[i],b[j],b[jj])) 179 { 180 flag=true; 181 break; 182 } 183 } 184 if(flag) 185 break; 186 } 187 if(flag) 188 break; 189 } 190 if(flag) 191 break; 192 } 193 if(flag) 194 printf("intersect "); 195 else 196 printf("disjoint "); 197 198 } 199 } 200 return 0; 201 }