LA2797 Monster Trap

题意

PDF

分析

可以考虑建图，跑迷宫。

然后以线段端点，原点，和无穷大点建图，有边的条件是两点连线和墙没有交点。

但是对两个线段的交点处理就会有问题，所以把线段延长。另外还需要判断延长后在墙上，舍去这些端点。

时间复杂度\(O(T n^3)\)

代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<ctime>
#include<cstring>
#define rg register
#define il inline
#define co const
template<class T>il T read()
{
	rg T data=0;
	rg int w=1;
	rg char ch=getchar();
	while(!isdigit(ch))
	{
		if(ch=='-')
			w=-1;
		ch=getchar();
	}
	while(isdigit(ch))
	{
		data=data*10+ch-'0';
		ch=getchar();
	}
	return data*w;
}
template<class T>T read(T&x)
{
	return x=read<T>();
}
using namespace std;
typedef long long ll;

co double eps=1e-12;
double dcmp(double x)
{
	return fabs(x)<eps?0:(x<0?-1:1);
}

struct Point
{
	double x,y;

	Point(double x=0,double y=0)
	:x(x),y(y){}

	bool operator<(co Point&p)co
	{
		return x<p.x||(x==p.x&&y<p.y);
	}

	bool operator==(co Point&p)co
	{
		return x==p.x&&y==p.y;
	}
};
typedef Point Vector;

Vector operator+(co Vector&A,co Vector&B)
{
	return Vector(A.x+B.x,A.y+B.y);
}

Vector operator-(co Point&A,co Point&B)
{
	return Vector(A.x-B.x,A.y-B.y);
}

Vector operator*(co Point&A,double v)
{
	return Vector(A.x*v,A.y*v);
}

Vector operator/(co Point&A,double v)
{
	return Vector(A.x/v,A.y/v);
}

double Cross(co Vector&A,co Vector&B)
{
	return A.x*B.y-A.y*B.x;
}

double Dot(co Vector&A,co Vector&B)
{
	return A.x*B.x+A.y*B.y;
}

double Length(co Vector&A)
{
	return sqrt(Dot(A,A));
}

bool SegmentProperIntersection(co Point&a1,co Point&a2,co Point&b1,co 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(co Point&p,co Point&a1,co Point&a2)
{
	return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;
}

co int MAXV=202;
int V;
int G[MAXV][MAXV],vis[MAXV];

bool dfs(int u)
{
	if(u==1)
		return 1;
	vis[u]=1;
	for(int v=0;v<V;++v)
		if(G[u][v]&&!vis[v]&&dfs(v))
			return 1;
	return 0;
}

co int MAXN=100;
int n;
Point p1[MAXN],p2[MAXN];

bool OnAnySegment(Point p)
{
	for(int i=0;i<n;++i)
		if(OnSegment(p,p1[i],p2[i]))
			return 1;
	return 0;
}

bool IntersectWithAnySegment(Point a,Point b)
{
	for(int i=0;i<n;++i)
		if(SegmentProperIntersection(a,b,p1[i],p2[i]))
			return 1;
	return 0;
}

bool find_path()
{
	vector<Point>vertices;
	vertices.push_back(Point(0,0));
	vertices.push_back(Point(1e5,1e5));
	for(int i=0;i<n;++i)
	{
		if(!OnAnySegment(p1[i]))
			vertices.push_back(p1[i]);
		if(!OnAnySegment(p2[i]))
			vertices.push_back(p2[i]);
	}
	V=vertices.size();
	for(int i=0;i<V;++i)
		fill(G[i],G[i]+V,0);
	fill(vis,vis+V,0);
	for(int i=0;i<V;++i)
		for(int j=i+1;j<V;++j)
			if(!IntersectWithAnySegment(vertices[i],vertices[j]))
				G[i][j]=G[j][i]=1;
	return dfs(0);
}

int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	while(read(n))
	{
		for(int i=0;i<n;++i)
		{
			int x1,y1,x2,y2;
			read(x1),read(y1),read(x2),read(y2);
			Point a=Point(x1,y1);
			Point b=Point(x2,y2);
			Vector v=b-a;
			v=v/Length(v);
			p1[i]=a-v*1e-6;
			p2[i]=b+v*1e-6;
		}
		puts(find_path()?"no":"yes");
	}
	return 0;
}