poj1696

Posted 2021-02-11 mxang

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

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <iostream>
 4 #include <vector>
 5 #include <algorithm>
 6 #include <cstring>
 7 using namespace std;
 8 typedef double db;
 9 const db eps=1e-6;
10 const db pi=acos(-1);
11 int sign(db k){
12     if (k>eps) return 1; else if (k<-eps) return -1; return 0;
13 }
14 int cmp(db k1,db k2){return sign(k1-k2);}
15 struct point{
16     db x,y;int id;
17     point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
18     point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
19     point operator * (db k1) const{return (point){x*k1,y*k1};}
20     point operator / (db k1) const{return (point){x/k1,y/k1};}
21     db abs(){return sqrt(x*x+y*y);}
22 };
23 db cross(point k1,point k2){ return k1.x*k2.y-k2.x*k1.y;}
24 db dot(point k1,point k2){ return k1.x*k2.x+k1.y*k2.y;}
25 int t,n;
26 point p[55];
27 bool cmp2(point a,point b){
28     return a.y<b.y;
29 }
30 vector <point>ans;
31 int vis[55];
32 int main(){
33     ios::sync_with_stdio(false);
34     cin>>t;
35     while (t--){
36         cin>>n;
37         for(int i=1;i<=n;i++){
38             cin>>p[i].id>>p[i].x>>p[i].y;
39         }
40         sort(p+1,p+1+n,cmp2);
41         p[0].x=0,p[0].y=p[1].y;
42         ans.push_back(p[0]);
43         ans.push_back(p[1]);
44         vis[1]=1;
45         for(int i=2;i<=n;i++){
46             int m=ans.size();
47             point now = ans[m-1]-ans[m-2];//之前的向量，找最小左偏。
48 
49             int id=-1;
50             db mn=-2;
51             for(int j=1;j<=n;j++){
52                 if(vis[j])continue;
53                 point tmp = p[j]-ans[m-1];
54                 db c = dot(tmp,now)/tmp.abs()/now.abs();
55                 if(cmp(c,mn)==1){
56                     //printf("%lf %lf
",tmp.x,tmp.y);
57                     mn = c;
58                     id=j;
59                 }
60             }
61             if(id==-1)break;
62             else {
63                 ans.push_back(p[id]);
64                 vis[id]=1;
65             }
66         }
67         cout<<n<<" ";
68         for(int i=1;i<ans.size();i++){
69             cout<<ans[i].id;
70             if(i!=ans.size()-1)
71                 cout<<‘ ‘;
72         }
73         cout<<endl;
74         ans.clear();
75         memset(vis,0, sizeof(vis));
76     }
77 }

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