lightoj1107 How Cow

思路：此题就是简单叉积运用，判断点在规则图形内，比如三角形，平行四边形等。

// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <climits>
using namespace std;
// #define DEBUG
#ifdef DEBUG
#define debug(...) printf( __VA_ARGS__ )
#else
#define debug(...)
#endif
#define CLR(x) memset(x, 0,sizeof x)
#define MEM(x,y) memset(x, y,sizeof x)
#define pk push_back
template<class T> inline T Get_Max(const T&a,const T&b){return a < b?b:a;}
template<class T> inline T Get_Min(const T&a,const T&b){return a < b?a:b;}
// typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
const double eps = 1e-10;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct Point{
	double x, y;
	Point(){}
	Point(double _x, double _y){
		x = _x;
		y = _y;
	}
	Point operator + (const Point& rhs)const{
		return Point(x + rhs.x, y + rhs.y);
	}
	Point operator - (const Point& rhs)const{
		return Point(x - rhs.x, y - rhs.y);
	}
	double operator ^ (const Point& rhs)const{
		return (x * rhs.y - y * rhs.x);
	}
	double operator * (const Point& rhs)const{
		return (x * rhs.x + y * rhs.y);
	}
	Point operator * (const double Num)const{
		return Point(x * Num,y * Num);
	}
	friend ostream& operator << (ostream& output,const Point& rhs){
		output << "(" << rhs.x << "," << rhs.y << ")";
		return output;
	}
};
typedef Point Vector;
/*向量的模*/
inline double Length(const Vector& A){//Get the length of vector A;
	return sqrt(A * A);//sqrt(x*x+y*y);
}
/*向量夹角*/
inline double Angle(const Vector& A,const Point& B){
	return acos(A * B/Length(A)/Length(B));
}
/*向量A旋转rad弧度*/
inline Point Rotate(const Vector& A, double rad){
	return Vector(A.x*cos(rad) - A.y*sin(rad),A.x*sin(rad) + A.y*cos(rad));
}
Point LL,LU,RU,RL;
double S;
bool check(Point O){
	double x = O^RL;
	double s1 = fabs((O - LL)^(RL - O)) / 2.0;
	double s2 = fabs((RL - O)^(RU - O)) / 2.0;
	double s3 = fabs((RU - O)^(LU - O)) / 2.0;
	double s4 = fabs((LU - O)^(LL - O)) / 2.0;
	if (s1 < eps || s2 < eps || s3 < eps || s4 < eps) return false;
	return (s1 + s2 + s3 + s4 <= S + eps);
}
int main()
{	// ios::sync_with_stdio(false);
	// freopen("in.txt","r",stdin);
	// freopen("out.txt","w",stdout);
	int t, icase = 0, n;
	cin >> t;
	while(t--){
		cin >> LL.x >> LL.y >> RU.x >> RU.y;
		LU = Point(LL.x, RU.y);
		RL = Point(RU.x, LL.y);
		S = (RU.x - LL.x) * (RU.y - LL.y);
		cin >> n;
		int ans = 0;
		printf("Case %d:\n", ++icase);
		while(n--){
			double x, y;
			cin >> x >> y;
			if (check(Point(x,y))) puts("Yes");
			else puts("No");
		}
	}
	return 0;
}