不正方形（四个点构成一个凸四边形）

链接：http://oj.saikr.com/problem/ADPC2-C
题意：两黄个点得在两个红点连线的两边且红点也得在黄点连线的两边

#include<stdio.h>
#include<vector>
using namespace std;
typedef long long ll;
struct node{
	ll x,y;
}A[255],B[255];
vector<int>v1,v2;
int main()
{
	int N,M;
	scanf("%d %d",&N,&M);
	for(int i=0;i<N;i++)
		scanf("%lld %lld",&A[i].x,&A[i].y);
	for(int i=0;i<M;i++)
		scanf("%lld %lld",&B[i].x,&B[i].y);
	
	bool Flag=false;
	for(int i1=0;i1<N;i1++)
	{
		for(int i2=i1+1;i2<N;i2++)
		{
			/*TL
			for(int j1=0;j1<M;j1++)
			{
				for(int j2=j1+1;j2<M;j2++)
				{
					ll a=A[i2].y-A[i1].y,b=A[i1].x-A[i2].x;
					ll c=A[i1].y*(A[i2].x-A[i1].x)+A[i1].x*(A[i1].y-A[i2].y);
					int f1=0,f2=0;
					if(A[i1].y==A[i2].y)
					{
						if(B[j1].y<A[i1].y) f1=-1;
						else if(B[j1].y>A[i1].y) f1=1;
						if(B[j2].y<A[i2].y) f2=-1;
						else if(B[j2].y>A[i2].y) f2=1;
					}
					else if(A[i1].x==A[i2].x)
					{
						if(B[j1].x<A[i1].x) f1=-1;
						else if(B[j1].x>A[i1].x) f1=1;
						if(B[j2].x<A[i2].x) f2=-1;
						else if(B[j2].x>A[i2].x) f2=1;
					}
					else
					{
						if(a*B[j1].x+b*B[j1].y+c<0) f1=-1;
						else if(a*B[j1].x+b*B[j1].y+c>0) f1=1;
						if(a*B[j2].x+b*B[j2].y+c<0) f2=-1;
						else if(a*B[j2].x+b*B[j2].y+c>0) f2=1;
					}
					
					if(f1!=0&&f1!=f2)
					{
						a=B[j2].y-B[j1].y,b=B[j1].x-B[j2].x;
						c=B[j1].y*(B[j2].x-B[j1].x)+B[j1].x*(B[j1].y-B[j2].y);
						if(B[j1].x==B[j2].x)
						{
							if(A[i1].x<B[j1].x) f1=-1;
							else if(A[i1].x>B[j1].x) f1=1;
							else f1=0;
							if(A[i2].x<B[j2].x) f2=-1;
							else if(A[i2].x>B[j2].x) f2=1;
							else f2=0;
						}
						else if(B[j1].y==B[j2].y)
						{
							if(A[i1].y<B[j1].y) f1=-1;
							else if(A[i1].y>B[j1].y) f1=1;
							else f1=0;
							if(A[i2].y<B[j2].y) f2=-1;
							else if(A[i2].y>B[j2].y) f2=1;
							else f2=0;
						}
						else
						{
							if(a*A[i1].x+b*A[i1].y+c<0) f1=-1;
							else if(a*A[i1].x+b*A[i1].y+c>0) f1=1;
							else f1=0;
							if(a*A[i2].x+b*A[i2].y+c<0) f2=-1;
							else if(a*A[i2].x+b*A[i2].y+c>0) f2=1;
							else f2=0;
						}
						
						if(f1!=0&&f1!=f2)
							Flag=true,i1=i2=N,j1=j2=M;
					}
				}
			}
			*/
			v1.clear(),v2.clear();//别忘了claer
			for(int i=0;i<M;i++)
			{
				ll a=A[i2].y-A[i1].y,b=A[i1].x-A[i2].x;
				ll c=A[i1].y*(A[i2].x-A[i1].x)+A[i1].x*(A[i1].y-A[i2].y);
				if(A[i1].y==A[i2].y)
				{
					if(B[i].y<A[i1].y) v1.push_back(i);
					else if(B[i].y>A[i1].y) v2.push_back(i);
				}
				else if(A[i1].x==A[i2].x)
				{
					if(B[i].x<A[i1].x) v1.push_back(i);
					else if(B[i].x>A[i1].x) v2.push_back(i);
				}
				else
				{
					if(a*B[i].x+b*B[i].y+c<0) v1.push_back(i);
					else if(a*B[i].x+b*B[i].y+c>0) v2.push_back(i);
				}
			}
			
			for(int j=0;j<v1.size();j++)
			{
				for(int k=0;k<v2.size();k++)
				{
					int j1=v1[j],j2=v2[k];
					ll a=B[j2].y-B[j1].y,b=B[j1].x-B[j2].x;
					ll c=B[j1].y*(B[j2].x-B[j1].x)+B[j1].x*(B[j1].y-B[j2].y);
					int f1=0,f2=0;
					if(B[j1].x==B[j2].x)
					{
						if(A[i1].x<B[j1].x) f1=-1;
						else if(A[i1].x>B[j1].x) f1=1;
						if(A[i2].x<B[j2].x) f2=-1;
						else if(A[i2].x>B[j2].x) f2=1;
					}
					else if(B[j1].y==B[j2].y)
					{
						if(A[i1].y<B[j1].y) f1=-1;
						else if(A[i1].y>B[j1].y) f1=1;
						if(A[i2].y<B[j2].y) f2=-1;
						else if(A[i2].y>B[j2].y) f2=1;
					}
					else
					{
						if(a*A[i1].x+b*A[i1].y+c<0) f1=-1;
						else if(a*A[i1].x+b*A[i1].y+c>0) f1=1;
						if(a*A[i2].x+b*A[i2].y+c<0) f2=-1;
						else if(a*A[i2].x+b*A[i2].y+c>0) f2=1;
					}
					
					if(f1!=0&&f1!=f2)
						Flag=true,i1=N,i2=N,j=v1.size(),k=v2.size();//break
				}
			}
		}
	}
	if(Flag==true) printf("YES");
	else printf("NO");
	
	
	return 0;
}