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欧拉定理题意: 给你N 个点,按顺序一笔画完连成一个多边形
求这个平面被分为多少个区间
欧拉定理 : 平面上边为 n ,点为 c 则 区间为 n + 2 - c;
思路: 先扫,两两线段的交点,存下来,有可能会有重复点, 用STL unique 去重
在把每个点判断是否在线段上,有,则会多出一条线段(不包括端点的点)
这样就得出结果了 O(n * n ) /// 这题还有点卡精度, 1e-11 WA, 1e-9 A了…….
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<set> #include<cmath> using namespace std; const int maxn = 300 +31; const double eps = 1e-9; struct Point { double x, y; Point(double x = 0,double y = 0) : x(x), y(y) {} }; int dcmp(double x) { if(fabs(x) < eps) return 0 ; else return x < 0 ? -1 : 1; } 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) { return a.x < b.x || (a.x==b.x && a.y < b.y); } 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; } Point GetLineIntersection (Point P, Vector v, Point Q, Vector w) { ///求直线焦点 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) { /// 判断两线段是否相交 (不包括端点) 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 (Point p, Point a1, Point a2) { /// 判断点是否在线段上(不包括端点) return dcmp(Cross(a1-p,a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0; } Point P[maxn], V[maxn*maxn]; int main() { int n,kase = 1; while( cin >> n && n ) { for(int i = 0; i < n; ++i) { cin >> P[i].x >> P[i].y; V[i] = P[i]; } n--; int c = n; for(int i = 0; i < n; ++i) { for(int j = i+1; j < n; ++j) if(SegmentProperIntersection(P[i],P[i+1],P[j],P[j+1])) V[c++] = GetLineIntersection(P[i],P[i+1]-P[i],P[j],P[j+1]-P[j]); } sort(V,V+c); c = unique(V,V+c) - V;///顶点个数 int e = n;///边 for(int i = 0; i < c; ++i) { for(int j = 0; j < n; ++j) if(OnSegment(V[i],P[j],P[j+1])) e++; } int sum = e+2-c; //if(sum == 1) printf("Case %d: There are %d piece.\n",kase++,sum); printf("Case %d: There are %d pieces.\n",kase++,sum); } return 0; }
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原文地址:http://www.cnblogs.com/aoxuets/p/4742409.html