学步园

大意：在Dukeswood这块土地上生活着一个富有的农庄主和他的几个孩子。在他临终时，他想把他的土地分给他的孩子。他有许多农场，每个农场都是一块矩形土地。他在农场地图上划上一些直线将矩形分成若干块。当他划直线时，他总是从矩形边界上的某一点划到另一个矩形边界上的点，这条线的结束点将成为下一条线的起始点。他划线时从不会让任三线共点。题目链接

思路：设f(n)为前n条输入线段将矩形分成区域的个数。

边界：f(1) = 2，假设已经产生了n-1条线段，新线段为l，它和已有的n-1条线段有T(n)个交点，这些交点将l分成了T(n)+1条线段，线段将所在区域一分为2，所以新增了T(n)+1个区域，所以f(n) = f(n-1)+T(n)+1，递推可知：f(L) = f(1) + T+L-1 = T+L+1，其中L为总的线段相交的交点数目。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
using namespace std;

const double eps = 1e-10;

struct Point
{
	double x, y;
	Point(double x = 0, double y = 0) : x(x), y(y) { }
};

typedef Point Vector;

Vector operator + (Vector A, Vector B) { return Vector(A.x+B.x, A.y+B.y); }

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

Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }

Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }

bool operator < (const Point& a, const Point &b)
{
	if(a.x != b.x) return a.x < b.x;
	return a.y < b.y;
}

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

bool operator == (const Point& a, const Point &b)
{
	return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
}

double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }

double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
double Area2(Point A, Point B, Point C) { return Cross(B-A, C-A); }

Vector Rotate(Vector A, double rad)
{
	return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad));
}

Point GetIntersection(Point P, Vector v, Point Q, Vector w)
{
	Vector u = P-Q;
	double t = Cross(w, u) / Cross(v, w);
	return P+v*t;
}

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

double PolygonArea(Point* p, int n)
{
	double area = 0;
	for(int i = 1; i < n-1; i++)
		area += Cross(p[i]-p[0], p[i+1]-p[0]);
	return area;
}

Point read_point()
{
	Point A;
	scanf("%lf%lf", &A.x, &A.y);
	return A;
}

int w, h;
int n;

const int maxn = 60;

Point p[maxn];

int read_case()
{
	scanf("%d%d", &w, &h);
	if(!w) return 0;
	scanf("%d", &n);
	for(int i = 0; i <= n; i++) p[i] = read_point();
	return 1;
}

void solve()
{
	int tot = 0;
	for(int i = 0; i <= n-1; i++)
	{
		for(int j = i+1; j <= n-1; j++)
		{
			Point a1 = p[i], a2 = p[i+1];
			Point b1 = p[j], b2 = p[j+1];
			if(SegmentProperIntersection(a1, a2, b1, b2)) tot++;
		}
	}
	int ans = n+tot+1;
	printf("%d\n", ans);
}

int main()
{
	while(read_case())
	{
		solve();
	}
	return 0;
}