UVA 1342 - That Nice Euler Circuit(计算几何+欧拉定理)

大白上的例题

思路：首先要知道欧拉定理， 顶点数V，边数E，面数F，那么有V + F - E = 2

那么剩下就是根据已有的图形，计算出有多少个顶点和多少条边，就能计算出面数了

于是暴力计算几何搞搞即可

代码：

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;struct Point {    double x, y;    Point() {}    Point(double x, double y) {        this->x = x;        this->y = y;    }    void read() {        scanf("%lf%lf", &x, &y);    }};typedef Point Vector;Vector operator + (Vector A, Vector B) {    return Vector(A.x + B.x, A.y + B.y);}Vector operator - (Vector A, Vector 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);}const double eps = 1e-8;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;} //点积double Length(Vector A) {return sqrt(Dot(A, A));} //向量的模double Angle(Vector A, Vector B) {return acos(Dot(A, B) / Length(A) / Length(B));} //向量夹角double Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;} //叉积double Area2(Point A, Point B, Point C) {return Cross(B - A, C - A);} //有向面积//向量旋转Vector Rotate(Vector A, double rad) {    return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));}//计算两直线交点，平行，重合要先判断Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {    Vector u = P - Q;    double t = Cross(w, u) / Cross(v, w);    return P + v * t;}//点到直线距离double DistanceToLine(Point P, Point A, Point B) {    Vector v1 = B - A, v2 = P - A;    return fabs(Cross(v1, v2)) / Length(v1);}//点到线段距离double DistanceToSegment(Point P, Point A, Point B) {    if (A == B) return Length(P - A);    Vector v1 = B - A, v2 = P - A, v3 = P - B;    if (dcmp(Dot(v1, v2)) < 0) return Length(v2);    else if (dcmp(Dot(v1, v3)) > 0) return Length(v3);    else return fabs(Cross(v1, v2)) / Length(v1);}//点在直线上的投影点Point GetLineProjection(Point P, Point A, Point B) {    Vector v = B - A;    return A + v * (Dot(v, P - A) / Dot(v, v));}//线段相交判定(规范相交)bool SegmentProperIntersection(Point a1, Point a2, Point b1, 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;}//判断点在线段上, 不包含端点bool OnSegment(Point p, Point a1, Point a2) {    return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;}//n边形的面积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 / 2;}const int N = 305;int n;Point p[N], v[N * N];int main() {    int cas = 0;    while (~scanf("%d", &n) && n) {        for (int i = 0; i < n; i++) {            p[i].read();            v[i] = p[i];        }        n--;        int vn = 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[vn++] = GetLineIntersection(p[i], p[i + 1] - p[i], p[j], p[j + 1] - p[j]);                }            }        }        sort(v, v + vn);        vn = unique(v, v + vn) - v;        for (int i = 0; i < vn; 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", ++cas, e - vn + 2);    }    return 0;}