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-6;
const db pi=acos(-1);
int sign(db k){
    if (k>eps) return 1; else if (k<-eps) return -1; return 0;
}
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[55];
bool cmp2(point a,point b){
    return a.y<b.y;
}
vector <point>ans;
int vis[55];
int main(){
    ios::sync_with_stdio(false);
    cin>>t;
    while (t--){
        cin>>n;
        for(int i=1;i<=n;i++){
            cin>>p[i].id>>p[i].x>>p[i].y;
        }
        sort(p+1,p+1+n,cmp2);
        p[0].x=0,p[0].y=p[1].y;
        ans.push_back(p[0]);
        ans.push_back(p[1]);
        vis[1]=1;
        for(int i=2;i<=n;i++){
            int m=ans.size();
            point now = ans[m-1]-ans[m-2];//之前的向量，找最小左偏。

            int id=-1;
            db mn=-2;
            for(int j=1;j<=n;j++){
                if(vis[j])continue;
                point tmp = p[j]-ans[m-1];
                db c = dot(tmp,now)/tmp.abs()/now.abs();
                if(cmp(c,mn)==1){
                    //printf("%lf %lf
",tmp.x,tmp.y);
                    mn = c;
                    id=j;
                }
            }
            if(id==-1)break;
            else {
                ans.push_back(p[id]);
                vis[id]=1;
            }
        }
        cout<<n<<" ";
        for(int i=1;i<ans.size();i++){
            cout<<ans[i].id;
            if(i!=ans.size()-1)
                cout<<' ';
        }
        cout<<endl;
        ans.clear();
        memset(vis,0, sizeof(vis));
    }
}