LA 3236 That Nice Euler Circuit（欧拉定理）

That Nice Euler Circuit

Timelimit:3.000 seconds

Little Joey invented a scrabble machine that he called Euler, after the great mathematician. In his primary school Joey heard about the nice story of how Euler started the study about graphs. The problem in that
story was - let me remind you - to draw a graph on a paper without lifting your pen, and finally return to the original position. Euler proved that you could do this if and only if the (planar) graph you created has the following two properties: (1) The graph
is connected; and (2) Every vertex in the graph has even degree.

Joey‘s Euler machine works exactly like this. The device consists of a pencil touching the paper, and a control center issuing a sequence of instructions. The paper can be viewed as the infinite two-dimensional plane; that means you do not need to worry about
if the pencil will ever go off the boundary.

In the beginning, the Euler machine will issue an instruction of the form (X0, Y0) which moves the pencil to some starting position (X0, Y0).
Each subsequent instruction is also of the form (X‘, Y‘), which means to move the pencil from the previous position to the new position (X‘, Y‘), thus draw a line segment
on the paper. You can be sure that the new position is different from the previous position for each instruction. At last, the Euler machine will always issue an instruction that move the pencil back to the starting position(X0, Y0).
In addition, the Euler machine will definitely not draw any lines that overlay other lines already drawn. However, the lines may intersect.

After all the instructions are issued, there will be a nice picture on Joey‘s paper. You see, since the pencil is never lifted from the paper, the picture can be viewed as an Euler circuit.

Your job is to count how many pieces (connected areas) are created on the paper by those lines drawn by Euler.

Input

There are no more than 25 test cases. Ease case starts with a line containing an integer N4,
which is the number of instructions in the test case. The following N pairs of integers give the instructions and appear on a single line separated by single spaces. The first pair is the first instruction that gives the
coordinates of the starting position. You may assume there are no more than 300 instructions in each test case, and all the integer coordinates are in the range (-300, 300). The input is terminated when N is 0.

Output

For each test case there will be one output line in the format

Case x: There are w pieces.,

where x is the serial number starting from 1.

Note: The figures below illustrate the two sample input cases.

Sample Input

5
0 0 0 1 1 1 1 0 0 0
7
1 1 1 5 2 1 2 5 5 1 3 5 1 1
0

Sample Output

Case 1: There are 2 pieces.
Case 2: There are 5 pieces.

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<algorithm>
using namespace std;

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);
}

const double eps = 1e-10;
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;

        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.\n", ++cas, ans);
    }
    return 0;
}