UVALive - 3263 That Nice Euler Circuit (几何蓝书)

￡神魔★判官ぃ 2022-05-27 08:53 127阅读 0赞

#include<cstdio>
    #include<cmath>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    const double ep=1e-10;
    const int maxn=3e2+10;
    
    struct Point {
    	double x, y;
    	Point(){}
    	Point(double x, double y): x(x), y(y){}
    	bool operator < (const Point A) const{ return x<A.x || (x==A.x&&y<A.y);}
    	
    };
    typedef Point Vector;
    
    int dcmp(double x){
    	if(fabs(x) < ep) return 0; else return x<0? -1 : 1;
    }
    
    Point operator + (Point A, Vector B) { return Point(A.x+B.x, A.y+B.y); }
    Point 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 (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; }
    
    bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2){
    	double c1=Cross(a2-a1, b1-a1), c2=Cross(a2-a1, b2-a1);
    	double c3=Cross(b2-b1, a1-b1), c4=Cross(b2-b1, a2-b1);
    	return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
    }
    
    Point GetLineIntersection(Point a1, Vector v1, Point a2, Vector v2){
    	Vector v=a1-a2;
    	double t=Cross(v2, v)/Cross(v1, v2);
    	return a1+v1*t;
    }
    
    bool OnSegment(Point p, Point a, Point b){
    	return dcmp(Cross(a-p, b-p))==0 && dcmp(Dot(a-p, b-p))<0;
    }
    
    int main() {
    	int n, cas=0;
    	Point  p[maxn], v[maxn*maxn];
    	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 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]))
    					v[c++]=GetLineIntersection(p[i], p[i+1]-p[i], p[j], p[j+1]-p[j]);
    		
    		sort(v, v+c);
    		c=unique(v, v+c)-v;//一定要加上原先的点再去重，因为原先的的端点可能落在原先的线段上，当再有边穿过时，并没有产生新边。
    		for(int i=0; i<c; i++)
    			for(int j=0; j<n; j++)
    				if(OnSegment(v[i], p[j], p[j+1])) e++;
    		
    		printf("Case %d: There are %d pieces.\n", ++cas, 2+e-c);
    	}
    	return 0;
    }