pku 2284 That Nice Euler Circuit

题意：

给你ｎ个点第ｎ个点保证与第０个点相交，然后求这ｎ个点组成的图形可以把整个平面分成几个面

思路：

这里的解题关键是知道关于多面体的欧拉定理　

多面体：

设v为顶点数，e为棱数，f是面数，则
v-e+f=2-2p
p为欧拉示性数，例如
p=0 的多面体叫第零类多面体
p=1 的多面体叫第一类多面体

这里满足的是零类多面体，我们只要求出该图形的　点v,边e即可。 怎么求点v呢？ 两部分一部分是原来的n-1个顶点，然后是交出来的，我们只要判断线段相交求直线交点即可，然偶可能会摇头重复的交点去掉，求边的话我们只要求出一个规范相交的点肯定会增加一条边，枚举点然后判断 点是否在线段上（除了端点），然偶求解即可。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val) memset(arr, val, sizeof(arr))

#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("din.txt", "r", stdin)
#define Write() freopen("d.out", "w", stdout)
#define ll unsigned long long
#define keyTree (chd[chd[root][1]][0])

#define M 100007
#define N 317

using namespace std;

const int inf = 0x7f7f7f7f;
const int mod = ;

struct Point
{
    double x,y;
    Point(double tx = ,double ty = ) : x(tx),y(ty){}
};
typedef Point Vtor;

Vtor operator + (Vtor A,Vtor B) { return Vtor(A.x + B.x,A.y + B.y); }
Vtor operator - (Point A,Point B) { return Vtor(A.x - B.x,A.y - B.y); }
Vtor operator * (Vtor A,double p) { return Vtor(A.x*p,A.y*p); }
Vtor operator / (Vtor A,double p) { return Vtor(A.x/p,A.y/p); }
const double eps = 1e-;
int dcmp(double x) { if (fabs(x) < eps) return ; else return x > ?  : -; }
bool operator < (Point A,Point B) { return A.x < B.x || (A.x == B.x && A.y < B.y); }
bool operator == (Point A,Point B) { return dcmp(A.x - B.x) ==  && dcmp(A.y - B.y) ==  ; }

double Dot(Vtor A,Vtor B) { return A.x*B.x + A.y*B.y; }
double Length(Vtor A) { return sqrt(Dot(A,A)); }
double Angle(Vtor A,Vtor B) { return acos(Dot(A,B)/Length(A)/Length(B)); }
double Cross(Vtor A,Vtor B) { return A.x*B.y - A.y*B.x; }
double Area2(Point A,Point B,Point C) { return Cross(B - A,C - A); }
Vtor Rotate(Vtor A,double rad) { return Vtor(A.x*cos(rad) - A.y*sin(rad),A.x*sin(rad) + A.y*cos(rad)); }
/*>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>*/

bool SegmentPI(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*c2) <  && dcmp(c3*c4) < ;

}
bool OnSegment(Point p,Point a1,Point a2)
{
    return dcmp(Cross(a2 - p, a1 - p)) ==  && dcmp(Dot(a2 - p,a1 - p)) < ;
}
Point GetLineIn(Point P,Vtor v,Point Q,Vtor w)
{
    Vtor u = P - Q;
    double t = Cross(w,u)/Cross(v,w);
    return P + v*t;
}
Point P[N],v[N*N];
int n;

int main()
{

    Read();
    int Ei,Vi;
    int cas = ;
    while (~scanf("%d",&n))
    {
        if (!n) break;
        Vi = ;
        for (int i = ; i < n; ++i)
        scanf("%lf%lf",&P[i].x,&P[i].y),v[Vi++] = P[i];

        Ei = --n;
        for (int i = ; i < n; ++i)
        {
            for (int j = i + ; j < n; ++j)
            {
                if (SegmentPI(P[i],P[i + ],P[j],P[j + ]))
                {
                    v[Vi++] = GetLineIn(P[i],P[i + ] - P[i],P[j],P[j + ] - P[j]);
                }
            }
        }
        sort(v, v + Vi);
        Vi = unique(v,v + Vi) - v;
        for (int i = ; i < Vi; ++i)
        {
            for (int j = ; j < n; ++j)
            {
                if (OnSegment(v[i],P[j],P[j + ])) Ei++;
            }
        }
//        printf("%d %d\n",Vi,Ei);
        printf("Case %d: There are %d pieces.\n",cas++,Ei - Vi + );
    }

    return ;
}