POJ2079 旋转卡壳 凸包内最大三角形

先求个凸包，再求每个点的对踵点。假设I的对踵点是J，那I+1的对踵点必然在J+1后，所以可以用2个指针向后推。

利用叉积来求三角形的面积来找对踵点。

这题要注意的是，数据里既有3点共线，还有重点。

我的代码：

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#define MAXN 50010using namespace std;struct point{    int x,y;    point(int _x=0,int _y=0) : x(_x),y(_y) {}    void input() { scanf("%d%d",&x,&y); }};point operator-(const point p1,const point p2){  return point(p1.x-p2.x, p1.y-p2.y);}int operator*(const point p1,const point p2){  return p1.x*p2.y - p1.y*p2.x;}bool operator<(const point p1,const point p2){  if(p1.x != p2.x) return p1.x < p2.x;   else return p1.y < p2.y;}int n,tn;point a[MAXN],tu[MAXN];void init(){    int i,j,k,r,w;    for(i=1;i<=n;i++)       a[i].input();  }void tubao(){    int i,j,k,r,w,un,dn;    point dd[MAXN],du[MAXN];    sort(a+1,a+n+1);    un = dn = 2;    dd[1] = du[1] = a[1];    dd[2] = du[2] = a[2];        for(i=3;i<=n;i++)    {  for(;un > 1 && (du[un] - du[un-1]) * (a[i] - du[un]) > 0;un--);       for(;dn > 1 && (dd[dn] - dd[dn-1]) * (a[i] - dd[dn]) < 0;dn--);       du[++un] = a[i];       dd[++dn] = a[i];    }    tn = 0;    for(i=1;i<un;i++)  tu[++tn] = du[i];    for(i=dn;i>1;i--)  tu[++tn] = dd[i];}int dis(point p1, point p2){    return (p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y);}int myabs(int x){    if(x < 0) x = -x;     return x;}void solve(){    int i,j,k,r,w,old=0;    int ans=0;    tubao();    tu[0] = tu[tn];    tu[tn+1] = tu[1];    for(i=1,j=1;i<=tn;i++)    {  old = j;       while(  myabs( (tu[i+1]-tu[i]) * (tu[j]-tu[i]) )<=              myabs( (tu[i+1]-tu[i]) * (tu[j+1]-tu[i]) ))        {   ans = max(ans, dis(tu[i],tu[j]) );           j = j % tn + 1;           if( j== old) break;        }         ans = max(ans, dis(tu[i],tu[j]) );    }    printf("%d\n",ans);}int main(){    freopen("p2187.in","r",stdin);    while(scanf("%d",&n) != EOF)    {  init();       solve();    }    return 0;}