uvalive 3263 That Nice Euler Circuit

题意：平面上有一个包含n个端点的一笔画，第n个端点总是和第一个端点重合，因此团史一条闭合曲线。组成一笔画的线段可以相交，但是不会部分重叠。求这些线段将平面分成多少部分（包括封闭区域和无限大区域）。

分析：若是直接找出所有区域，或非常麻烦，而且容易出错。但用欧拉定理可以将问题进行转化，使解法变容易。

欧拉定理：设平面图的顶点数、边数和面数分别为V，E，F，则V+F-E=2。

这样，只需求出顶点数V和边数E，就可以求出F=E+2-V。

设平面图的结点由两部分组成，即原来的结点和新增的结点。由于可能出现三线共点，需要删除重复的点。

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<memory.h>
#include<cstdlib>
#include<vector>
#define clc(a,b) memset(a,b,sizeof(a))
#define LL long long int
using namespace std;
const int inf=0x3f3f3f3f;
const double eps = 1e-10;
const int N = 300 + 5;

struct Point
{
    double x, y;
    Point(double x = 0, double y = 0) : x(x), y(y) { }
};

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);
}

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;
}

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)
{
    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)
{
    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)
{
    return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;
}

Point P[N], V[N*N];

int main()
{
    int n, cas = 0;
    while(~scanf("%d",&n) && n)
    {
        for(int i = 0; i < n; i++)
        {
            scanf("%lf%lf", &P[i].x, &P[i].y);
            V[i] = P[i];
        }
        n--;
        int vcnt = n, ecnt = 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[vcnt++] = GetLineIntersection(P[i], P[i+1]-P[i], P[j], P[j+1]-P[j]);
            }
        sort(V, V+vcnt);
        vcnt = unique(V, V+vcnt) - V;//去掉相邻元素中重复的，使用前先排序
        for(int i = 0; i < vcnt; i++)
            for(int j = 0; j < n; j++)
                if(OnSegment(V[i], P[j], P[j+1]))
                    ecnt++;
        int ans = ecnt + 2 - vcnt;
        printf("Case %d: There are %d pieces.
", ++cas, ans);
    }
    return 0;
}