Squares

【題目鏈接】Squares

【題目類型】旋轉卡殼

&題解：

聽著算法名字,感覺挺難,仔細一看之后,發現其實很簡單,就是依靠所構成三角行面積來快速的找對踵點,就可以省去很多的復雜度了.旋轉的復雜度是O(n),之后還有計算每條邊對應的對踵點復雜度平均大約O(n/2)在實際中也許可以更快

&代碼：

#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個三角形面積相等

				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;

}

總結

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