FZU 1015 土地划分

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;}