2015长春 HDU 5531 Rebuild

题意：n个顶点组成的多边形能否形成正多边形？

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <cmath>
  4 #include <cstdlib>
  5 #include <iostream>
  6 #include <algorithm>
  7 #include <queue>
  8 #include <map>
  9 #include <vector>
 10 using namespace std;
 11 const int maxn=111;
 12 const int maxnode=4000010;
 13 typedef long long LL;
 14 struct Point
 15 {
 16     int x,y;
 17     Point (int x=0,int y=0):x(x),y(y) {}
 18
 19 } p[maxn];
 20 typedef Point Vector;
 21 Vector operator+(Vector A,Vector B)
 22 {
 23     return Vector(A.x+B.x,A.y+B.y);
 24 }
 25
 26 Vector operator-(Point A,Point B)
 27 {
 28     return Vector(A.x-B.x,A.y-B.y);
 29 }
 30 Vector operator*(Vector A,int p)
 31 {
 32     return Vector(A.x*p,A.y*p);
 33 }
 34
 35 Vector operator /(Vector A,int p)
 36 {
 37     return Vector(A.x/p,A.y/p);
 38 }
 39
 40 bool operator <(const Point&a,const Point &b)
 41 {
 42     return a.x<b.x||(a.x==b.x&&a.y<b.y);
 43 }
 44
 45 int Cross(Vector A,Vector B)
 46 {
 47     return A.x*B.y-A.y*B.x;
 48 }
 49
 50 int ConvexHull(Point *p,int n,Point *ch)
 51 {
 52     sort(p,p+n);
 53     int m=0;
 54     for(int i=0; i<n; i++)
 55     {
 56         while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)
 57             m--;
 58         ch[m++]=p[i];
 59     }
 60     int k=m;
 61     for(int i=n-2; i>=0; i--)
 62     {
 63         while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;
 64         ch[m++]=p[i];
 65     }
 66     if(n>1)
 67         m--;
 68     return m;
 69 }
 70 int n;
 71 Point ans[maxn];
 72 LL  dis(LL X1,LL Y1,LL X2,LL Y2)
 73 {
 74     return (X1-X2)*(X1-X2)+(Y1-Y2)*(Y1-Y2);
 75 }
 76 int main()
 77 {
 78     freopen("in.txt","r",stdin);
 79     int t ;
 80     scanf("%d",&t);
 81     while(t--)
 82     {
 83         scanf("%d",&n);
 84         for(int i=0; i<n; i++)
 85             scanf("%d%d",&p[i].x,&p[i].y);
 86         int m=ConvexHull(p,n,ans);
 87         if(m!=n)
 88         {
 89             printf("NO\n");
 90             continue;
 91         }
 92         bool flag=true;
 93         LL len=dis(ans[0].x,ans[0].y,ans[m-1].x,ans[m-1].y);
 94
 95         for(int i=1; i<m; i++)
 96         {
 97             if(len!=dis(ans[i].x,ans[i].y,ans[i-1].x,ans[i-1].y))
 98             {
 99                 flag=false;
100                 break;
101             }
102         }
103         if(flag)printf("YES\n");
104         else printf("NO\n");
105
106     }
107     return 0;
108 }