2019-2020 ICPC Asia Taipei-Hsinchu Regional Contest

2019-2020 ICPC Asia Taipei-Hsinchu Regional Contest

L

题意：给N个点（N $\le$ 4096），求四边形最大面积（只要有四个点即可，可以退化成三角形或直线，注意重复点的情况）
题解：建凸包，旋转卡壳跑每个点的对踵点。对于一对对踵点，枚举其他点，找出直线两侧面积最大、最小值。如果两侧都有点，用两侧的最大值相加更新答案；否则用有点的那一侧的最大值减最小值更新答案。
O(N²)

#include
#include
#include
#include
#include
#include
#define LL long long
using namespace std;
const int N=5010;
const LL inf=2e18;
struct NODE{
	LL x,y;
	NODE(){}
	NODE (LL a,LL b){ x=a; y=b; }
	NODE friend operator - (NODE a,NODE b)
	{	return NODE(a.x-b.x,a.y-b.y);	} 
}node[N];
int que[N];
int stop,n;
LL X(NODE a,NODE b)
{
	return a.x*b.y-a.y*b.x;
} 
LL dist(NODE a,NODE b)
{
	return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
int cmp(NODE a,NODE b)
{
	NODE x=a-node[1];
	NODE y=b-node[1];
	LL s=X(x,y);
	return s>0||(s==0&&dist(node[1],a)<=dist(node[1],b));
}
LL S(int a,int b,int c)
{
	NODE x=node[b]-node[a];
	NODE y=node[c]-node[a];
	return X(x,y);
}
LL cal(int a,int b)
{
	LL mx1,mx2,mn1,mn2;
	mx1=mx2=-1;
	mn1=mn2=inf;
	for(int i=1;i<=n;i++)
	if(i!=a&&i!=b)
		{
			LL s=S(a,b,i);
			if(s>0)
				mx1=max(mx1,s),mn1=min(mn1,s);
			else
				mx2=max(mx2,-s),mn2=min(mn2,-s);
		}
	if(mx1==-1)
		return mx2-mn2;
	if(mx2==-1)
		return mx1-mn1;
	return mx1+mx2;
} 
int main()
{
	int T;
	scanf("%d",&T);
	while(T)
	{
		T--;
		LL ans=0;
		scanf("%d",&n);
		for(int i=1;i<=n;i++)
			cin>>node[i].x>>node[i].y;
		int k=1;
		for(int i=1;i<=n;i++)
			if(node[i].y<node[k].y||(node[i].y==node[k].y&&node[i].x<node[k].x))
				k=i;
		swap(node[k].x,node[1].x);	swap(node[k].y,node[1].y);
		sort(node+2,node+n+1,cmp);
		que[0]=1;	que[1]=2;
		stop=1;
		for(int i=3;i<=n;i++)
		{
			while(stop>=1&&S(que[stop-1],que[stop],i)<=0)
				stop--;
			que[++stop]=i;
		}
		if(stop<=1)
		{
			printf("0\n");
			continue;
		}
		int j=1;
		for(int i=0;i<stop;i++)
		{
			while(abs(S(que[i],que[(i+1)%stop],que[j%stop]))<abs(S(que[i],que[(i+1)%stop],que[(j+1)%stop])))
				j++;
			ans=max(ans,cal(que[i],que[j%stop]));
		}
		if(ans&1)
			cout<<ans/2<<".5"<<endl;
		else
			cout<<ans/2<<endl;
	}
	return 0;
 }