poj1696

没看题解，搜了一下都是什么叉积凸包，根本没有必要用吧。。

显然这个题我们找夹角就可以了，根据高中的公式 a·b=|a|*|b|*cos<a,b>

所以用点积找一个 cos<a,b> 最小的角就可以了。

一发ac稳得一批

 #include <cstdio>
 #include <cmath>
 #include <iostream>
 #include <vector>
 #include <algorithm>
 #include <cstring>
 using namespace std;
 typedef double db;
 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;int id;
     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 * (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);}
 };
 db cross(point k1,point k2){ return k1.x*k2.y-k2.x*k1.y;}
 db dot(point k1,point k2){ return k1.x*k2.x+k1.y*k2.y;}
 int t,n;
 point p[];
 bool cmp2(point a,point b){
     return a.y<b.y;
 }
 vector <point>ans;
 int vis[];
 int main(){
     ios::sync_with_stdio(false);
     cin>>t;
     while (t--){
         cin>>n;
         for(int i=;i<=n;i++){
             cin>>p[i].id>>p[i].x>>p[i].y;
         }
         sort(p+,p++n,cmp2);
         p[].x=,p[].y=p[].y;
         ans.push_back(p[]);
         ans.push_back(p[]);
         vis[]=;
         for(int i=;i<=n;i++){
             int m=ans.size();
             point now = ans[m-]-ans[m-];//之前的向量，找最小左偏。

             int id=-;
             db mn=-;
             for(int j=;j<=n;j++){
                 if(vis[j])continue;
                 point tmp = p[j]-ans[m-];
                 db c = dot(tmp,now)/tmp.abs()/now.abs();
                 if(cmp(c,mn)==){
                     //printf("%lf %lf\n",tmp.x,tmp.y);
                     mn = c;
                     id=j;
                 }
             }
             if(id==-)break;
             else {
                 ans.push_back(p[id]);
                 vis[id]=;
             }
         }
         cout<<n<<" ";
         for(int i=;i<ans.size();i++){
             cout<<ans[i].id;
             if(i!=ans.size()-)
                 cout<<' ';
         }
         cout<<endl;
         ans.clear();
         memset(vis,, sizeof(vis));
     }
 }