UVALive 3263 That Nice Euler Circuit
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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,但是不会部分重叠,给出各个顶点的坐标,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)
解题思路:用欧拉定理将问题进行转化。欧拉定理:设平面内的顶点数、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。该平面图的结点有两部分组成,及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
代码:
#include <iostream>#include <algorithm>#include <string>#include <cstring>#include <cstdio>#include <cmath>#include <vector>#include <queue>using namespace std;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 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-A.y*B.x;}double Area2(Point A,Point B,Point C){ return Cross(B-A,C-A);}Vector Rotate(Vector A,double rad){ return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}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;}const int maxn=310;Point P[maxn],V[maxn*maxn];int main(){ int n; int 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]; } int c=n,e=n-1;//原来的顶点数和边数 for(int i=0; i<n-1; i++) { for(int j=i+1; j<n-1; j++) { if(SegmentProperIntersection(P[i],P[i+1],P[j],P[j+1]))//两线段相交 { V[c++]=GetLineIntersection(P[i],P[i+1]-P[i],P[j],P[j+1]-P[j]);//交点 } } } sort(V,V+c); c=unique(V,V+c)-V;//去重点 for(int i=0; i<c; i++) { for(int j=0; j<n-1; j++) { if(Onsegment(V[i],P[j],P[j+1]))e++;//点在边上,边数就增加了 } } printf("Case %d: There are %d pieces.\n",++cas,e+2-c); } return 0;}
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