●POJ 2284 That Nice Euler Circuit

Posted 2020-10-20 *ZJ

题链：

http://poj.org/problem?id=2284

题解：

计算几何，平面图的欧拉定理

欧拉定理：设平面图的定点数为v，边数为e，面数为f，则有 v+f-e=2

即 f=e-v+2

所以$N^2$求出所以线段的交点，并去重，

然后再计算出最后共有多少边，（判断点是否在线段上，是的话则e++）

总的复杂度 $O(N^3)$

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAXN 350
using namespace std;
const double eps=1e-8;
struct Point{
	double x,y;
	Point(double _x=0,double _y=0):x(_x),y(_y){}
};
typedef Point Vector;
int sign(double x){
	if(-eps<=x&&x<=eps) return 0;
	return x<0?-1:1;
}
bool operator < (const Point &A,const Point &B){return sign(A.x-B.x)<0||(sign(A.x-B.x)==0&&sign(A.y-B.y)<0);}
bool operator == (const Point &A,const Point &B){return sign(A.x-B.x)==0&&sign(A.y-B.y)==0;}
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);}
double operator ^ (Vector A,Vector B){return A.x*B.y-A.y*B.x;}
double operator * (Vector A,Vector B){return A.x*B.x+A.y*B.y;}
Point D[MAXN],V[MAXN*MAXN];
int N;
bool SPI(Point a1,Point a2,Point b1,Point b2){//Segment_Proper_Intersection
	static double c1,c2,c3,c4;
	c1=(a2-a1)^(b1-a1); c2=(a2-a1)^(b2-a1);
	c3=(b2-b1)^(a1-b1); c4=(b2-b1)^(a2-b1);
	return sign(c1*c2)<0&&sign(c3*c4)<0;
}
bool OS(Point P,Point a1,Point a2){//On_Segment
	return sign((P-a1)^(P-a2))==0&&sign((P-a1)*(P-a2))<0;
}
Point GLI(Point P,Vector v,Point Q,Vector w){//Get_Line_Intersection
	static Vector u; u=P-Q;
	return P+v*((w^u)/(v^w));
}
int main(){
	int Case=0,v,e;
	while(scanf("%d",&N)&&N){
		for(int i=1;i<=N;i++) 
			scanf("%lf%lf",&D[i].x,&D[i].y),V[i]=D[i];
		N--; v=N; e=N;
		for(int i=1;i<=N;i++)
			for(int j=1;j<i;j++)
				if(SPI(D[j],D[j+1],D[i],D[i+1]))
					V[++v]=GLI(D[j],D[j+1]-D[j],D[i],D[i+1]-D[i]);
		sort(V+1,V+v+1);
		v=unique(V+1,V+v+1)-V-1;
		for(int i=1;i<=v;i++)
			for(int j=1;j<=N;j++)
				if(OS(V[i],D[j],D[j+1])) e++;
		printf("Case %d: There are %d pieces.\n",++Case,e-v+2);
	}
	return 0;
}