POJ 2284-That Nice Euler Circuit(计算几何_欧拉定理求平面被分成的区域数)

That Nice Euler Circuit

Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2284

Appoint description: System Crawler  (2015-05-16)

Description

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 N >= 4, 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

50 0 0 1 1 1 1 0 0 071 1 1 5 2 1 2 5 5 1 3 5 1 10

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，图案是一条闭合曲线，组成一笔画的线段可以相交，但是不会不分重叠，求这些线段将平面分为多少部分。

思路：用欧拉定理。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

该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，所以需要删除重复的点。

#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <sstream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>using namespace std;typedef long long LL;const int inf=0x3f3f3f3f;const double eps=1e-10;const double pi= acos(-1.0);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) { //向量+向量=向量，向量+点=点    return Vector(A.x+B.x,A.y+B.y);}Vector operator -(Point A,Point B) { //点-点=向量    return Vector(A.x-B.x,A.y-B.y);}Vector operator *(Vector A,double p) { //向量*数=向量    return Vector(A.x*p,A.y*p);}Vector operator /(Vector A,double p) { //向量/数=向量    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) { //求点积    return A.x*B.x+A.y*B.y;}double Length(Vector A) { //利用点积求向量长度    return sqrt(Dot(A,A));}double Angle(Vector A,Vector B) { //利用点积求夹角    return acos(Dot(A,B)/Length(A)/Length(B));}double Cross(Vector A,Vector B) { // 叉积    return A.x*B.y-A.y*B.x;}double Area(Point A,Point B,Point C) {    return Cross(B-A,C-A);}Vector Rotate(Vector A,double rad) { //向量的逆时针旋转    return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}//调用前确保两条直线相交。当且仅当Cross(v,w)非零。Point GetLineIntersection(Point P,Vector v, Point Q,Vector w) { //计算两条直线P+tv和Q+tw的交点    Vector u=P-Q;    double t=Cross(w,u)/Cross(v,w);    return P+v*t;}//线段规范相交的充要条件是:每条线段的两个端点都在另一条线段的两侧(两侧指叉积的符号不同)bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) {//线段相交判定    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;}// 判断一个点是否在一条线段(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;}const int maxn=310;Point P[maxn],V[maxn*maxn];int main(){    int n, i,j;    int icase = 1;    while(~scanf("%d",&n)) {            if(!n) break;        for(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(i = 0; i < n; i++)            for(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(i = 0; i < Vcnt; i++)            for(j = 0; j < n; j++)                if(OnSegment(V[i], P[j], P[j+1]))                    Ecnt++;        printf("Case %d: There are %d pieces.\n", icase++, Ecnt+2-Vcnt);    }    return 0;}