UVAL 4728 Squares(旋转卡壳)

Squares

【题目链接】Squares

【题目类型】旋转卡壳

&题解：

听着算法名字,感觉挺难,仔细一看之后,发现其实很简单,就是依靠所构成三角行面积来快速的找对踵点,就可以省去很多的复杂度了.旋转的复杂度是O(n),之后还有计算每条边对应的对踵点复杂度平均大约O(n/2)在实际中也许可以更快

&代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;

struct Point {
	int x,y;
	Point(int x=0,int y=0): x(x),y(y) {}
};
typedef Point Vector;
Vector operator - (const Vector& A,const Vector& B) {return Vector(A.x-B.x , A.y-B.y);}
bool operator == (const Point& a,const Point& b) {return a.x==b.x&&a.y==b.y;}
bool operator < (const Point& a,const Point& b) {return a.x<b.x||(a.x==b.x&&a.y<b.y);}
int Cross(Vector A,Vector B) {return A.x*B.y - A.y*B.x;}
int Dist2(const Point& a,const Point& b) {return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}

vector<Point> ConvexHull(vector<Point> p) {
	sort(p.begin(),p.end());
	p.erase(unique(p.begin(),p.end()) , p.end());
	int n=p.size(), m=0;
	vector<Point> ch(n+1);
	for(int i=0; i<n; i++) {
		while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-1])<=0) m--;
		ch[m++]=p[i];
	}
	int k=m;
	for(int i=n-2; i>=0; i--) {
		while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-1])<=0) m--;
		ch[m++]=p[i];
	}
	if(m>1) m--;
	ch.resize(m);
	return ch;
}

int diameter2(vector<Point>& points) {
	vector<Point> p=ConvexHull(points);
	int n=p.size();
	if(n==1) return 0;
	if(n==2) return Dist2(p[0] , p[1]);
	p.push_back(p[0]);
	int ans=0;
	for(int u=0, v=1; u<n; u++) {
		for(;;) {
			int diff = Cross(p[u+1]-p[u] , p[v+1]-p[v]);
			if( diff<=0 ) {
				ans=max(ans , Dist2(p[u],p[v]));
				//2个三角形面积相等
				if(diff==0) ans=max(ans, Dist2(p[u], p[v+1]));
				break;
			}
			v=(v+1)%n;
		}
	}
	return ans;
}

int main() {
	freopen("E:1.in","r",stdin);
	int T;
	scanf("%d",&T);
	while(T--) {
		int n;
		scanf("%d" ,&n);
		vector<Point> po;
		for(int i=0; i<n; i++) {
			int x,y,w;
			scanf("%d%d%d",&x,&y,&w);
			po.push_back(Point(x,y));
			po.push_back(Point(x+w,y));
			po.push_back(Point(x,y+w));
			po.push_back(Point(x+w,y+w));
		}
		printf("%d\n", diameter2(po));
	}
	return 0;
}