That Nice Euler Circuit（LA3263+几何）
That Nice Euler Circuit
Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %lluDescription
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
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.
/********************************************************************************* 欧拉定理：设平面图的顶点数、边数和面数分别为V、E和F，则V+F-E=2 so...F=E+2-V; 该平面图的结点有原来的和新增结点构成，由于可能出现三线共点，需要删除重复点 *********************************************************************************/ #include<cstdio> #include<cmath> #include<algorithm> #define PI acos(-1.0) using namespace std; struct Point { double x,y; Point(double x=0,double y=0):x(x),y(y){} }; typedef Point Vector; //向量+向量=向量; 向量+点=点 Vector operator + (Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);} //点-点=向量 Vector operator - (Point A,Point B){return Vector(A.x-B.x,A.y-B.y);} //向量*数=向量 Vector operator * (Vector A,double p){return Vector(A.x*p,A.y*p);} //向量/数=向量 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 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-B.x*A.y;} double Area2(Point A,Point B,Point C){return Cross(B-A,C-A);} /*******两直线交点*******/ //调用前确保两条直线P+tv和Q+tv有唯一交点，当且仅当Cross(v,w)非0; Point GetLineIntersection(Point P,Vector v,Point Q,Vector w) { Vector u=P-Q; if(Cross(v,w)) { double t=Cross(w,u)/Cross(v,w);//精度高的时候，考虑自定义分数类 return P+v*t; } // else // return ; } /************************ 线段相交判定（规范相交） ************************/ bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2) { double c1=Cross(a2-a1,b1-a1); double c2=Cross(a2-a1,b2-a1); double c3=Cross(b2-b1,a1-b1); double c4=Cross(b2-b1,a2-b1); return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0; } /**如果允许在端点处相交：如果c1和c2都是0，表示共线，如果c1和c2不都是0，则表示某个端点在另一条直线上**/ 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 mmax=310; Point P[mmax],V[mmax*mmax]; Point read_point(Point &P) { scanf("%lf%lf",&P.x,&P.y); return P; } int main() { int n; int ck=1; while(scanf("%d",&n),n) { for(int i=0;i<n;i++) { P[i]=read_point(P[i]); V[i]=P[i]; } n--; int c=n,e=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]))//严格相交 { V[c++]=GetLineIntersection(P[i],P[i+1]-P[i],P[j],P[j+1]-P[j]);//交点 } } } // printf("c=%d\n",c); sort(V,V+c); c=unique(V,V+c)-V; // printf("%d=%d-%d\n",c,unique(V,V+c),V); for(int i=0;i<c;i++) { for(int j=0;j<n;j++) { if(Onsegment(V[i],P[j],P[j+1])) e++;//边数 } } printf("Case %d: There are %d pieces.\n",ck++,e+2-c); } return 0; } /* 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 7 1 1 1 5 2 1 2 5 5 1 3 9 1 1 0 */
