题目大意：一个平面区域有n条线段，问能否从（0，0）处到达无穷远处（不穿过任何线段）
分析：若两条线段有一个端点重合，这种情况是不能从端点重合处穿过的 的。因此对每个端点延长一点，就可以避免这个问题。
n*2个端点加上起始点跟终点，两两之间不穿过任何线段的为可行路径建图。
最后以（0，0）开始dfs，看能否到达无穷远点。
#include<iostream>
#include<vector>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std; const double eps = 1e-12;
double dcmp(double x)
{
if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1;
} struct Point {
double x, y;
Point(double x=0, double y=0):x(x),y(y) { }
}; typedef Point Vector; Vector operator + (const Point& A, const Point& B) { return Vector(A.x+B.x, A.y+B.y);}
Vector operator - (const Point& A, const Point& B) { return Vector(A.x-B.x, A.y-B.y);}
Vector operator * (const Point& A, double v) { return Vector(A.x*v, A.y*v);}
Vector operator / (const Point& A, double v) { return Vector(A.x/v, A.y/v);}
double Cross(const Vector& A, const Vector& B) { return A.x*B.y - A.y*B.x;}
double Dot(const Vector& A, const Vector& B) { return A.x*B.x + A.y*B.y;}
double Length(const Vector& A) { return sqrt(Dot(A,A));}
bool operator < (const Point& p1, const Point& p2) {
return p1.x < p2.x || (p1.x == p2.x && p1.y < p2.y);
}
bool operator == (const Point& p1, const Point& p2) {
return p1.x == p2.x && p1.y == p2.y;
} bool SegmentProperIntersection(const Point& a1, const Point& a2, const Point& b1, const 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(const Point& p, const Point& a1, const Point& a2) {
return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;
}
const int maxn = 100 + 5;
int n,V;
const int maxv = 200 + 5;
int G[maxv][maxv], vis[maxv];
Point p1[maxn], p2[maxn]; // 在任何一条线段的中间（在端点不算）
bool OnAnySegment(Point p) {
for(int i = 0; i < n; i++)
if(OnSegment(p, p1[i], p2[i])) return true;
return false;
} // 与任何一条线段规范相交
bool IntersectWithAnySegment(Point a, Point b) {
for(int i = 0; i < n; i++)
if(SegmentProperIntersection(a, b, p1[i], p2[i])) return true;
return false;
} bool dfs(int u)
{
if(u == 1) return true; // 1是终点
vis[u] = 1;
for(int v = 0; v < V; v++)
if(G[u][v] && !vis[v] && dfs(v)) return true;
return false;
} bool find_path()
{
// 构图
int i,j;
vector<Point> vertices;
vertices.push_back(Point(0, 0)); // 起点
vertices.push_back(Point(1e5, 1e5)); // 终点
for(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();
memset(G, 0, sizeof(G));
memset(vis, 0, sizeof(vis));
for(i = 0; i < V; i++)
for(j = i+1; j < V; j++)
if(!IntersectWithAnySegment(vertices[i], vertices[j]))
G[i][j] = G[j][i] = 1;
return dfs(0);
} int main()
{
double x1, y1, x2, y2;
int i;Vector v;
while(scanf("%d",&n),n)
{
for(i = 0; i < n; i++)
{
scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
Point a = Point(x1, y1);
Point b = Point(x2, y2);
v = b - a;
v = v / Length(v);
p1[i] = a - v * 1e-6;
p2[i] = b + v * 1e-6;
}
if(find_path()) printf("no\n");
else printf("yes\n");
}
return 0;
}