POJ 3348

旋 转卡壳水题。

直接使用旋转卡壳求距离。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>

using namespace std;
const int MAXN=50100;
struct point {
	int x,y;
}p[MAXN];
int n;
int st[MAXN],ans[MAXN];
int stop,cnt;

bool cmp(point A,point B){
	if(A.y<B.y) return true;
	else if(A.y==B.y){
		if(A.x<B.x) return true;
	}
	return false;
}

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

bool multi(point a,point b,point c){
	point p,q;
	p.x=a.x-c.x; p.y=a.y-c.y;
	q.x=b.x-c.x; q.y=b.y-c.y;
	return (p.x*q.y-q.x*p.y)>=0;
}

int TriangleArea(point a,point b,point c){
	point p,q;
	p.x=a.x-c.x; p.y=a.y-c.y;
	q.x=b.x-c.x; q.y=b.y-c.y;
	return abs(p.x*q.y-q.x*p.y);
}

void slove(){
	stop=cnt=0;
	st[stop++]=0; st[stop++]=1;
	for(int i=2;i<n;i++){
		while(stop>1&&multi(p[i],p[st[stop-1]],p[st[stop-2]])) stop--;
		st[stop++]=i;
	}
	for(int i=0;i<stop;i++)
	ans[cnt++]=st[i];
	stop=0; st[stop++]=n-1; st[stop++]=n-2;
	for(int i=n-3;i>=0;i--){
		while(stop>1&&multi(p[i],p[st[stop-1]],p[st[stop-2]])) stop--;
		st[stop++]=i;
	}
	for(int i=1;i<stop;i++)
	ans[cnt++]=st[i];
}

int for_dist(){
	int distant=0;
	int q=1;
	for(int i=0;i<cnt-1;i++){
		while(TriangleArea(p[ans[i]],p[ans[i+1]],p[ans[q+1]])>
		TriangleArea(p[ans[i]],p[ans[i+1]],p[ans[q]])) 
		q=(q+1)%cnt;
		distant=max(distant,max(max(dist(p[ans[i]],p[ans[q]]),dist(p[ans[i+1]],p[ans[q+1]])),dist(p[ans[i+1]],p[ans[q]])));
	}
	return distant;
}

int main(){
	while(scanf("%d",&n)!=EOF){
		for(int i=0;i<n;i++){
			scanf("%d%d",&p[i].x,&p[i].y);
		}
		sort(p,p+n,cmp);
		slove();
		printf("%d\n",for_dist());
	}
	return 0;
}