LA 3263 (欧拉定理）

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;
}