poj2187

题意：求凸包直径。

分析：用graham按纵坐标序，再用旋转卡壳法。

但是模板我并没有完全理解，为什么要用点到直线的距离呢

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
usingnamespace std;

#define maxn 50005

struct Xpoint
{
    int x, y;
} pnt[maxn], res[maxn];

int n;

bool mult(Xpoint sp, Xpoint ep, Xpoint op)
{
    return (sp.x - op.x) * (ep. y - op.y) >= (ep.x - op.x) * (sp.y - op.y);
}

booloperator<(const Xpoint &l, const Xpoint &r)
{
    return l.y < r. y || (l.y == r.y && l.x < r.x);
}

int cross(Xpoint a, Xpoint b, Xpoint o)
{
    return (a.x - o.x) * (b.y - o.y) - (b.x - o.x) * (a.y - o.y);
}

int graham()
{
    int i, len, top =1;
    sort(pnt, pnt + n);
    if (n ==0)
        return0;
    res[0] = pnt[0];
    if (n ==1)
        return1;
    res[1] = pnt[1];
    if (n ==2)
        return2;
    res[2] = pnt[2];
    for (i =2; i < n; i++)
    {
        while (top && mult(pnt[i], res[top], res[top -1]))
            top--;
        res[++top] = pnt[i];
    }
    len = top;
    res[++top] = pnt[n -2];
    for (i = n -3; i >=0; i--)
    {
        while (top != len && mult(pnt[i], res[top], res[top -1]))
            top--;
        res[++top] = pnt[i];
    }
    return top;
}

int dist2(Xpoint a, Xpoint b)
{
    return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}

int rotating_calipers(Xpoint *ch, int n)
{
    int q =1, ans =0;
    ch[n] = ch[0];
    for (int p =0; p < n; p++)
    {
        while (cross(ch[p +1], ch[q +1], ch[p]) > cross(ch[p +1], ch[q], ch[p]))
            q = (q +1) % n;
        ans = max(ans, max(dist2(ch[p], ch[q]), dist2(ch[p +1], ch[q +1])));
    }
    return ans;
}

int main()
{
    //freopen("t.txt", "r", stdin);
    scanf("%d", &n);
    for (int i =0; i < n; i++)
        scanf("%d%d", &pnt[i].x, &pnt[i].y);
    int count = graham();
    int ans = rotating_calipers(res, count);
    printf("%d\n", ans);
    return0;
}