LA 3263 That Nice Euler Circuit (点、面、边的关系)

欧拉定理：设平面图的顶点数为V,边数为E，面数为F ，则，V+F-E=2；

但要注意顶点和边的计数，特别的去除重复的点

#include<cstdio>#include<cmath>#include<algorithm>using namespace std;const double eps = 1e-10;int dcmp(double x){    if(fabs(x) < eps) return 0;    else return x < 0 ? -1 : 1;}struct Point{    double x, y;    Point(double x=0, double y=0):x(x),y(y) { }};typedef Point Vector;Vector operator + (const Vector& A, const Vector& B){    return Vector(A.x+B.x, A.y+B.y);}Vector operator - (const Point& A, const Point& B){    return Vector(A.x-B.x, A.y-B.y);}Vector operator * (const 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(const Vector& A, const Vector& B){    return A.x*B.x + A.y*B.y;}double Cross(const Vector& A, const Vector& B){    return A.x*B.y - A.y*B.x;}Point GetLineIntersection(const Point& P, const Vector& v, const Point& Q, const Vector& w){    Vector u = P-Q;    double t = Cross(w, u) / Cross(v, w);    return P+v*t;}bool SegmentProperIntersection(const Point& a1, const Point& a2, const Point& b1, const Point& b2){    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(const Point& p, const Point& a1, const Point& a2){    return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;}const int maxn = 300 + 10;Point P[maxn], V[maxn*maxn];int main(){    int n, kase = 0;    while(scanf("%d", &n) == 1 && n)    {        for(int i = 0; i < n; i++)        {            scanf("%lf%lf", &P[i].x, &P[i].y);            V[i] = P[i];        }        n--;        int c = n, e = 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;  //去除重复的点        for(int i = 0; i < c; i++)            for(int j = 0; j < n; j++)                if(OnSegment(V[i], P[j], P[j+1])) e++;        printf("Case %d: There are %d pieces.\n", ++kase, e+2-c);    }    return 0;}