学步园

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;
}