gym 101657 D

理论1A。  //没删debug的文件读入。。

傻逼题。

先求出来每条边两侧的三角形，然后枚举边，根据叉积判断三角形位置，建图，拓扑排序。

 #include <bits/stdc++.h>
 #define pii pair<int,int>
 using namespace std;
 typedef double db;
 const int N = 1e6+;
 const db eps = 1e-;
 const db pi = acos(-);
 int sign(db k){
     if (k>eps) return ; else if (k<-eps) return -; return ;
 }
 int cmp(db k1,db k2){return sign(k1-k2);}
 struct point{
     db x,y;
     point operator+(const point &k1)const { return point{k1.x+x,k1.y+y};}
     point operator-(const point &k1)const { return point{x-k1.x,y-k1.y};}
     point operator*(const db k1)const { return point{x*k1,y*k1};}
     point operator / (db k1) const{return (point){x/k1,y/k1};}
     db abs(){return sqrt(x*x+y*y);}
     bool operator<(const point k1)const {
         int a = cmp(x,k1.x);
         if (a==-) return ; else if (a==) return ; else return cmp(y,k1.y)==-;
     }
     bool operator== (const point k1)const {
         return x==k1.x&&y==k1.y;
     }
 };
 db cross(point k1,point k2){ return k1.x*k2.y-k1.y*k2.x;}
 db dot(point k1,point k2){ return k1.x*k2.x+k1.y*k2.y;}
 struct line{
     point p[];//x小的是p[0]
     line(point k1,point k2){
         if(k1<k2)
             p[]=k1,p[]=k2;
         else
             p[]=k2,p[]=k1;
     }
     point &operator[](int k){ return p[k];}
     bool operator <(const line &k1)const {
         if(p[]==k1.p[])
             return p[]<k1.p[];
         return p[]<k1.p[];
     }
 };
 struct Tri{
     point p[],o;
     point &operator[](int k){ return p[k];}
 }tri[N];
 map<line,int> mp;
 vector<line> L;
 int cnt = ;
 vector<int> g[N*],f[N];
 int deg[N];
 int t,n;
 int main(){
     //freopen("awsl.in","r",stdin);
     scanf("%d",&t);
     while (t--){
         scanf("%d",&n);
         for(int i=;i<n;i++){
             tri[i].o.x=,tri[i].o.y=;
             for(int j=;j<;j++){
                 scanf("%lf%lf",&tri[i][j].x,&tri[i][j].y);
                 tri[i].o.x+=tri[i][j].x;
                 tri[i].o.y+=tri[i][j].y;
             }
             tri[i].o.x/=,tri[i].o.y/=;
             for(int j=;j<;j++){
                 line tmp = line(tri[i][j],tri[i][(j+)%]);
                 if(mp.count(tmp)){
                     g[mp[tmp]].push_back(i);
                 } else{
                     mp[tmp]=cnt;
                     L.push_back(tmp);
                     g[cnt].push_back(i);
                     cnt++;
                 }
             }
         }
         for(int i=;i<cnt;i++){
             if(g[i].size()<)continue;
             int s1 = g[i][],s2 = g[i][];
             line tmp = L[i];
             db t = cross(tmp[]-tmp[],tri[s1].o-tmp[]);
             if(sign(t)>){//s1在上面
                 f[s1].push_back(s2);
                 deg[s2]++;
             } else{
                 f[s2].push_back(s1);
                 deg[s1]++;
             }
         }
         queue<int> q;
         for(int i=;i<n;i++){
             if(deg[i]==)
                 q.push(i);
         }
         vector<int> ans;
         while (!q.empty()){
             int t = q.front();
             q.pop();
             ans.push_back(t+);
             for(int i=;i<f[t].size();i++){
                 deg[f[t][i]]--;
                 if(deg[f[t][i]]==)
                     q.push(f[t][i]);
             }
         }
         reverse(ans.begin(),ans.end());
         for(auto tmp:ans){
             printf("%d ",tmp);
         }
         printf("\n");
         for(int i=;i<n;i++){
             deg[i]=;
             f[i].clear();
         }
         for(int i=;i<cnt;i++)
             g[i].clear();
         L.clear();mp.clear();
         cnt=;
     }
 }