POJ2187-Beauty Contest-凸包

平面最远点对

由于点数为1e5，而整数点的情况下，凸包上点的个数为sqrt(M)，M为范围。

这样求出凸包之后n^2枚举维护距离就可以了

否则就用旋转卡壳。

这里用了挑战上的做法，比较简洁。

#include <cstdio>
#include <algorithm>
#define LL long long

using namespace std;

const int maxn = 5e4+10;
struct Point{
    int x,y;
    Point(int _x=0,int _y=0):x(_x),y(_y){}
    bool operator < (const Point &rhs) const{
        if(x == rhs.x) return y < rhs.y;
        else return x < rhs.x;
    }
    LL operator *(const Point &rhs) const{
        return (LL)x*rhs.x+(LL)y*rhs.y;
    }
    LL operator ^(const Point &rhs) const{
        return (LL)x*rhs.y - (LL)y*rhs.x;
    }
    Point operator -(const Point &rhs) const{
        return Point(x-rhs.x,y-rhs.y);
    }
}p[maxn],ch[maxn];
typedef Point Vector;

LL dist(Point a,Point b)
{
    return (a-b)*(a-b);
}

int ConvexHull(Point *pt,int n)
{
    sort(pt,pt+n);
    int k = 0;
    for(int i=0;i<n;i++)
    {
        while(k > 1 && ((ch[k-1]-ch[k-2])^(pt[i]-ch[k-1])) <= 0) k--;
        ch[k++] = pt[i];
    }
    for(int i=n-2,t = k;i>=0;i--)
    {
        while(k > t && ((ch[k-1]-ch[k-2])^(pt[i]-ch[k-1])) <= 0) k--;
        ch[k++] = pt[i];
    }
    return k;
}
int N;
int main()
{
    while(~scanf("%d",&N))
    {
        for(int i=0;i<N;i++)
        {
            scanf("%d%d",&p[i].x,&p[i].y);
        }
        int cnt = ConvexHull(p,N);
        LL ans = 0;
        for(int i=0;i<cnt;i++)
        {
            //printf("%d %d\n",ch[i].x,ch[i].y);
            for(int j=i+1;j<cnt;j++)
            {
                ans = max(ans,dist(ch[i],ch[j]));
            }
        }
        printf("%lld\n",ans);
    }
}