【POJ 2187】Beauty Contest（凸包直径、旋转卡壳）

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <algorithm>
 4 #define sqr(x) (x)*(x)
 5 #define  N 50001
 6 using namespace std;
 7 struct P{
 8     int x,y;    
 9 }p[N],q[N];
10 int n,ans,top;
11 int dis2(P a,P b){
12     return sqr(a.x-b.x)+sqr(a.y-b.y);
13 }
14 int xmult(P a,P b,P o){
15     return (a.x-o.x)*(b.y-o.y)-(a.y-o.y)*(b.x-o.x);
16 }
17 int cmp(P a,P b){
18     return xmult(a,b,p[1])>0||xmult(a,b,p[1])==0&&dis2(a,p[1])<dis2(b,p[1]);
19 }
20 void graham(){
21     sort(p+2,p+1+n,cmp);
22     q[1]=p[1],q[2]=p[2];
23     top=2;
24     for(int i=3;i<=n;i++){
25         while(xmult(q[top],p[i],q[top-1])<=0&&top>1)top--;
26         q[++top]=p[i];
27     }
28 }
29 void qiake(){
30     q[top+1]=q[1];
31     int now=2;
32     for(int i=1;i<=top;i++){
33         while(xmult(q[i+1],q[now],q[i])<xmult(q[i+1],q[now+1],q[i]))
34         {
35             now++;
36             if(now>top)now=1;
37         }
38         ans=max(ans,dis2(q[now],q[i]));
39     }
40 }
41 int main() {
42     scanf("%d",&n);
43     int t=1;
44     for(int i=1;i<=n;i++){
45         scanf("%d%d",&p[i].x,&p[i].y);
46         if(p[t].y>p[i].y||p[t].y==p[i].y&&p[t].x>p[i].x)t=i;
47     }
48     swap(p[1],p[t]);
49     graham();
50     qiake();
51     printf("%d",ans);
52     return 0;
53 }