uvalive 2797 Monster Trap

题意：给定一些线段障碍，判断怪物能不能逃离到无穷远处。

思路：从（0，0）点能否到无穷远处。用BFS搜索。那满足什么样的点符合要求，能加入到图中呢？

遍历每个点，显然一开始已经在某些线段上的点要删去。再判断，两点之间的连线是否与其他线段有交。有则删去。

这道题要注意如果两条线段重合，怎么办？延长每条线段，如果交点在线段内，不算。

 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<memory.h>
 #include<cstdlib>
 #include<vector>
 #define clc(a,b) memset(a,b,sizeof(a))
 #define LL long long int
 #define up(i,x,y) for(i=x;i<=y;i++)
 #define w(a) while(a)
 const double inf=0x3f3f3f3f;
 const int N = ;
 const double PI = acos(-1.0);
 using namespace std;
 const double eps = 1e-;
 
 double dcmp(double x)
 {
     if(fabs(x) < eps) return ;
     else return x <  ? - : ;
 }
 
 struct Point
 {
     double x, y;
     Point(double x=, double y=):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)< && dcmp(c3)*dcmp(c4)<;
 }
 
 bool OnSegment(const Point& p, const Point& a1, const Point& a2)
 {
     return dcmp(Cross(a1-p, a2-p)) ==  && dcmp(Dot(a1-p, a2-p)) < ;
 }
 
 const int maxv =  + ;
 int V;
 int G[maxv][maxv], vis[maxv];
 
 bool dfs(int u)
 {
     if(u == ) return true; // 1是终点
     vis[u] = ;
     for(int v = ; v < V; v++)
         if(G[u][v] && !vis[v] && dfs(v)) return true;
     return false;
 }
 
 const int maxn =  + ;
 int n;
 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 = ; 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 = ; i < n; i++)
         if(SegmentProperIntersection(a, b, p1[i], p2[i])) return true;
     return false;
 }
 
 bool find_path()
 {
     // 构图
     vector<Point> vertices;
     vertices.push_back(Point(, )); // 起点
     vertices.push_back(Point(1e5, 1e5)); // 终点
     for(int i = ; 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, , sizeof(G));
     memset(vis, , sizeof(vis));
     for(int i = ; i < V; i++)
         for(int j = i+; j < V; j++)
             if(!IntersectWithAnySegment(vertices[i], vertices[j]))
                 G[i][j] = G[j][i] = ;
     return dfs();
 }
 
 int main()
 {
     while(cin >> n && n)
     {
         for(int i = ; i < n; i++)
         {
             double x1, y1, x2, y2;
             cin >> x1 >> y1 >> x2 >> y2;
             Point a = Point(x1, y1);
             Point b = Point(x2, y2);
             Vector v = b - a;
             v = v / Length(v);
             p1[i] = a + v * 1e-;
             p2[i] = b + v * 1e-;
         }
         if(find_path()) cout << "no\n";
         else cout << "yes\n";
     }
     return ;
 }