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。
1 #include <bits/stdc++.h> 2 #define pii pair<int,int> 3 using namespace std; 4 typedef double db; 5 const int N = 1e6+6; 6 const db eps = 1e-6; 7 const db pi = acos(-1); 8 int sign(db k){ 9 if (k>eps) return 1; else if (k<-eps) return -1; return 0; 10 } 11 int cmp(db k1,db k2){return sign(k1-k2);} 12 struct point{ 13 db x,y; 14 point operator+(const point &k1)const { return point{k1.x+x,k1.y+y};} 15 point operator-(const point &k1)const { return point{x-k1.x,y-k1.y};} 16 point operator*(const db k1)const { return point{x*k1,y*k1};} 17 point operator / (db k1) const{return (point){x/k1,y/k1};} 18 db abs(){return sqrt(x*x+y*y);} 19 bool operator<(const point k1)const { 20 int a = cmp(x,k1.x); 21 if (a==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1; 22 } 23 bool operator== (const point k1)const { 24 return x==k1.x&&y==k1.y; 25 } 26 }; 27 db cross(point k1,point k2){ return k1.x*k2.y-k1.y*k2.x;} 28 db dot(point k1,point k2){ return k1.x*k2.x+k1.y*k2.y;} 29 struct line{ 30 point p[2];//x小的是p[0] 31 line(point k1,point k2){ 32 if(k1<k2) 33 p[0]=k1,p[1]=k2; 34 else 35 p[0]=k2,p[1]=k1; 36 } 37 point &operator[](int k){ return p[k];} 38 bool operator <(const line &k1)const { 39 if(p[0]==k1.p[0]) 40 return p[1]<k1.p[1]; 41 return p[0]<k1.p[0]; 42 } 43 }; 44 struct Tri{ 45 point p[3],o; 46 point &operator[](int k){ return p[k];} 47 }tri[N]; 48 map<line,int> mp; 49 vector<line> L; 50 int cnt = 0; 51 vector<int> g[N*3],f[N]; 52 int deg[N]; 53 int t,n; 54 int main(){ 55 //freopen("awsl.in","r",stdin); 56 scanf("%d",&t); 57 while (t--){ 58 scanf("%d",&n); 59 for(int i=0;i<n;i++){ 60 tri[i].o.x=0,tri[i].o.y=0; 61 for(int j=0;j<3;j++){ 62 scanf("%lf%lf",&tri[i][j].x,&tri[i][j].y); 63 tri[i].o.x+=tri[i][j].x; 64 tri[i].o.y+=tri[i][j].y; 65 } 66 tri[i].o.x/=3,tri[i].o.y/=3; 67 for(int j=0;j<3;j++){ 68 line tmp = line(tri[i][j],tri[i][(j+1)%3]); 69 if(mp.count(tmp)){ 70 g[mp[tmp]].push_back(i); 71 } else{ 72 mp[tmp]=cnt; 73 L.push_back(tmp); 74 g[cnt].push_back(i); 75 cnt++; 76 } 77 } 78 } 79 for(int i=0;i<cnt;i++){ 80 if(g[i].size()<2)continue; 81 int s1 = g[i][0],s2 = g[i][1]; 82 line tmp = L[i]; 83 db t = cross(tmp[1]-tmp[0],tri[s1].o-tmp[0]); 84 if(sign(t)>0){//s1在上面 85 f[s1].push_back(s2); 86 deg[s2]++; 87 } else{ 88 f[s2].push_back(s1); 89 deg[s1]++; 90 } 91 } 92 queue<int> q; 93 for(int i=0;i<n;i++){ 94 if(deg[i]==0) 95 q.push(i); 96 } 97 vector<int> ans; 98 while (!q.empty()){ 99 int t = q.front(); 100 q.pop(); 101 ans.push_back(t+1); 102 for(int i=0;i<f[t].size();i++){ 103 deg[f[t][i]]--; 104 if(deg[f[t][i]]==0) 105 q.push(f[t][i]); 106 } 107 } 108 reverse(ans.begin(),ans.end()); 109 for(auto tmp:ans){ 110 printf("%d ",tmp); 111 } 112 printf("\n"); 113 for(int i=0;i<n;i++){ 114 deg[i]=0; 115 f[i].clear(); 116 } 117 for(int i=0;i<cnt;i++) 118 g[i].clear(); 119 L.clear();mp.clear(); 120 cnt=0; 121 } 122 }
