1 /* 2 类型：多边形相交面积模板 3 */ 4 5 #include<cstdio> 6 #include<iostream> 7 #include<algorithm> 8 #include<cstring> 9 #include<cmath> 10 using namespace std; 11 #define maxn 510 12 const double eps=1E-8; 13 int sig(double d){ 14 return(d>eps)-(d<-eps); 15 } 16 struct Point{ 17 double x,y; Point(){} 18 Point(double x,double y):x(x),y(y){} 19 bool operator==(const Point&p)const{ 20 return sig(x-p.x)==0&&sig(y-p.y)==0; 21 } 22 }; 23 double cross(Point o,Point a,Point b){ 24 return(a.x-o.x)*(b.y-o.y)-(b.x-o.x)*(a.y-o.y); 25 } 26 double area(Point* ps,int n){ 27 ps[n]=ps[0]; 28 double res=0; 29 for(int i=0;i<n;i++){ 30 res+=ps[i].x*ps[i+1].y-ps[i].y*ps[i+1].x; 31 } 32 return res/2.0; 33 } 34 int lineCross(Point a,Point b,Point c,Point d,Point&p){ 35 double s1,s2; 36 s1=cross(a,b,c); 37 s2=cross(a,b,d); 38 if(sig(s1)==0&&sig(s2)==0) return 2; 39 if(sig(s2-s1)==0) return 0; 40 p.x=(c.x*s2-d.x*s1)/(s2-s1); 41 p.y=(c.y*s2-d.y*s1)/(s2-s1); 42 return 1; 43 } 44 //多边形切割 45 //用直线ab切割多边形p，切割后的在向量(a,b)的左侧，并原地保存切割结果 46 //如果退化为一个点，也会返回去,此时n为1 47 void polygon_cut(Point*p,int&n,Point a,Point b){ 48 static Point pp[maxn]; 49 int m=0;p[n]=p[0]; 50 for(int i=0;i<n;i++){ 51 if(sig(cross(a,b,p[i]))>0) pp[m++]=p[i]; 52 if(sig(cross(a,b,p[i]))!=sig(cross(a,b,p[i+1]))) 53 lineCross(a,b,p[i],p[i+1],pp[m++]); 54 } 55 n=0; 56 for(int i=0;i<m;i++) 57 if(!i||!(pp[i]==pp[i-1])) 58 p[n++]=pp[i]; 59 while(n>1&&p[n-1]==p[0])n--; 60 } 61 //---------------华丽的分隔线-----------------// 62 //返回三角形oab和三角形ocd的有向交面积,o是原点// 63 double intersectArea(Point a,Point b,Point c,Point d){ 64 Point o(0,0); 65 int s1=sig(cross(o,a,b)); 66 int s2=sig(cross(o,c,d)); 67 if(s1==0||s2==0)return 0.0;//退化，面积为0 68 if(s1==-1) swap(a,b); 69 if(s2==-1) swap(c,d); 70 Point p[10]={o,a,b}; 71 int n=3; 72 polygon_cut(p,n,o,c); 73 polygon_cut(p,n,c,d); 74 polygon_cut(p,n,d,o); 75 double res=fabs(area(p,n)); 76 if(s1*s2==-1) res=-res;return res; 77 } 78 //求两多边形的交面积 79 double intersectArea(Point*ps1,int n1,Point*ps2,int n2){ 80 if(area(ps1,n1)<0) reverse(ps1,ps1+n1); 81 if(area(ps2,n2)<0) reverse(ps2,ps2+n2); 82 ps1[n1]=ps1[0]; 83 ps2[n2]=ps2[0]; 84 double res=0; 85 for(int i=0;i<n1;i++){ 86 for(int j=0;j<n2;j++){ 87 res+=intersectArea(ps1[i],ps1[i+1],ps2[j],ps2[j+1]); 88 } 89 } 90 return res;//assumeresispositive! 91 } 92 //hdu-3060求两个任意简单多边形的并面积 93 Point ps1[maxn],ps2[maxn]; 94 int n1=3, n2=3; 95 int main(){ 96 int T; scanf("%d", &T); 97 while(T--) 98 { 99 for(int i=0;i<n1;i++) 100 scanf("%lf%lf",&ps1[i].x,&ps1[i].y); 101 for(int i=0;i<n2;i++) 102 scanf("%lf%lf",&ps2[i].x,&ps2[i].y); 103 double ans=intersectArea(ps1,n1,ps2,n2); 104 //ans=fabs(area(ps1,n1))+fabs(area(ps2,n2))-ans; //容斥，求并面积 105 double Area1=area(ps1, n1), Area2=area(ps2, n2); 106 if(ans>eps && min(Area1,Area2)-ans<eps) puts("contain") ; 107 else if(ans>eps && min(Area1,Area2)-ans>eps) puts("intersect"); 108 else 109 { 110 int mp=0; 111 for(int i=0; i<n1; ++i) 112 for(int j=0; j<n2; ++j) 113 if(ps1[i].x==ps2[j].x && ps1[i].y==ps2[j].y) 114 ++mp; 115 if(mp==2) puts("intersect"); 116 else puts("disjoint"); 117 } 118 } 119 return 0; 120 }