UVA10256 The Great Divide

题意

PDF

分析

就判断两个点集的凸包相离即可。

需要满足如下两个条件：

1. 两凸包上任意两条线段不相交。
2. 两凸包上任意一点不在另一凸包的内部。

时间复杂度\(O(T n m)\)

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<ctime>
#include<cstring>
#define rg register
#define il inline
#define co const
template<class T>il T read()
{
	rg T data=0;
	rg int w=1;
	rg char ch=getchar();
	while(!isdigit(ch))
	{
		if(ch=='-')
			w=-1;
		ch=getchar();
	}
	while(isdigit(ch))
	{
		data=data*10+ch-'0';
		ch=getchar();
	}
	return data*w;
}
template<class T>T read(T&x)
{
	return x=read<T>();
}
using namespace std;
typedef long long ll;

co double eps=1e-10;

double dcmp(double x)
{
	return fabs(x)<eps?0:(x<0?-1:1);
}

struct Point
{
	double x,y;
	
	Point(double x=0,double y=0)
	:x(x),y(y){}
	
	bool operator<(co Point&rhs)co
	{
		return x<rhs.x||(x==rhs.x&&y<rhs.y);
	}
	
	bool operator==(co Point&rhs)co
	{
		return x==rhs.x&&y==rhs.y;
	}
};
typedef Point Vector;

Vector operator-(co Vector&A,co Vector&B)
{
	return Vector(A.x-B.x,A.y-B.y);
}

double Dot(co Vector&A,co Vector&B)
{
	return A.x*B.x+A.y*B.y;
}

double Cross(co Vector&A,co Vector&B)
{
	return A.x*B.y-A.y*B.x;
}

bool SegmentProperIntersection(co Point&a1,co Point&a2,co Point&b1,co Point&b2)
{
	double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1),
		   c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);
	return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
}

bool OnSegment(co Point&p,co Point&a1,co Point&a2)
{
	return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;
}

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

int PointInPolygon(co Point&p,co vector<Point>&poly)
{
	int wn=0;
	int n=poly.size();
	for(int i=0;i<n;++i)
	{
		co Point&p1=poly[i];
		co Point&p2=poly[(i+1)%n];
		if(p1==p||p2==p||OnSegment(p,p1,p2))
			return -1;
		int k=dcmp(Cross(p2-p1,p-p1));
		int d1=dcmp(p1.y-p.y);
		int d2=dcmp(p2.y-p.y);
		if(k>0&&d1<=0&&d2>0)
			++wn;
		if(k<0&&d2<=0&&d1>0)
			--wn;
	}
	if(wn!=0)
		return 1;
	return 0;
}

bool ConvexPolygonDisjoint(co vector<Point>ch1,co vector<Point>&ch2)
{
	int c1=ch1.size();
	int c2=ch2.size();
	for(int i=0;i<c1;++i)
		if(PointInPolygon(ch1[i],ch2)!=0)
			return 0;
	for(int i=0;i<c2;++i)
		if(PointInPolygon(ch2[i],ch1)!=0)
			return 0;
	for(int i=0;i<c1;++i)
		for(int j=0;j<c2;++j)
			if(SegmentProperIntersection(ch1[i],ch1[(i+1)%c1],ch2[j],ch2[(j+1)%c2]))
				return 0;
	return 1;
}

int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	int n,m;
	while(read(n)|read(m))
	{
		vector<Point>P1,P2;
		for(int i=0;i<n;++i)
			P1.push_back(Point(read<int>(),read<int>()));
		for(int i=0;i<m;++i)
			P2.push_back(Point(read<int>(),read<int>()));
		if(ConvexPolygonDisjoint(ConvexHull(P1),ConvexHull(P2)))
			puts("Yes");
		else
			puts("No");
	}
	return 0;
}

来源：https://www.cnblogs.com/autoint/p/10162620.html