给你三根线段判段是否组成了A
条件，两条线段交于端点并且夹角不大于90，第三条线段端点在两条线段上并且划分的大小满足 大:小<4:1
注释很全。(主要是我记不清楚了，，好像过了一个多星期了)
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef long long db; 5 int inmid(db k1,db k2,db k3){return (k1-k3)*(k2-k3)<=0;}// k3 在 [k1,k2] 内 6 struct point{ 7 ll x,y; 8 point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};} 9 point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};} 10 point operator * (db k1) const{return (point){x*k1,y*k1};} 11 point operator / (db k1) const{return (point){x/k1,y/k1};} 12 int operator == (const point &k1) const{return x==k1.x&&y==k1.y;} 13 }; 14 int inmid(point k1,point k2,point k3){return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);} 15 ll cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;} 16 ll dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;} 17 struct line { 18 // p[0]->p[1] 19 point p[2]; 20 line(point k1, point k2) {p[0] = k1;p[1] = k2;} 21 point &operator[](int k) { return p[k]; } 22 }; 23 ll cross(line a,line b){return cross(a[1]-a[0],b[1]-b[0]);} 24 ll dot(line a,line b){return dot(a[1]-a[0],b[1]-b[0]);} 25 int t;point p[7]; 26 vector<line> l; 27 bool check(ll x,ll y){if(x*5<y)return 0;if(x*5>y*4)return 0;return 1;} 28 bool slove(line a,line b,line c){ 29 if(a[1]==b[1])swap(a[0],a[1]),swap(b[0],b[1]); 30 if(a[1]==b[0])swap(a[0],a[1]); 31 if(a[0]==b[1])swap(b[0],b[1]); 32 if(!(a[0]==b[0]))return 0;//判相交 33 if(cross(a,b)==0)return 0;//重合 34 if(dot(a,b)<0)return 0;//夹角 35 if(cross(a[1]-a[0],c[1]-a[0])==0&&cross(b[1]-b[0],c[0]-b[0])==0){} 36 else swap(c[0],c[1]); 37 if(!(cross(a[1]-a[0],c[1]-a[0])==0&&cross(b[1]-b[0],c[0]-b[0])==0))return 0;//共线 38 if(!inmid(a[0],a[1],c[1]))return 0;//在线段上 39 if(!inmid(b[0],b[1],c[0]))return 0; 40 //比例 41 if(!check(abs(a[0].x-c[1].x),abs(a[1].x-a[0].x)))return 0; 42 if(!check(abs(a[0].y-c[1].y),abs(a[1].y-a[0].y)))return 0; 43 if(!check(abs(b[0].x-c[0].x),abs(b[0].x-b[1].x)))return 0; 44 if(!check(abs(b[0].y-c[0].y),abs(b[0].y-b[1].y)))return 0; 45 return 1; 46 } 47 48 int main(){ 49 scanf("%d",&t); 50 while(t--){ 51 l.clear(); 52 for(int i=1;i<=6;i++)scanf("%lld%lld",&p[i].x,&p[i].y); 53 l.push_back({p[1],p[2]});l.push_back({p[3],p[4]});l.push_back({p[5],p[6]}); 54 if(slove(l[0],l[1],l[2]))cout<<"YES"<<endl; 55 else if(slove(l[0],l[2],l[1]))cout<<"YES"<<endl; 56 else if(slove(l[1],l[2],l[0]))cout<<"YES"<<endl; 57 else cout<<"NO"<<endl; 58 } 59 } 60 /** 61 1 62 0 0 0 6 63 0 6 2 -4 64 1 1 0 1 65 */