La 2797 (flat line diagram plsg) monster trap

Question:

There are N line segments on the plane, and the coordinates of the two endpoints of the N line segments are given at a time. Ask if the monster can escape from the coordinate origin to infinity. (If the two straight lines have at most one intersection and there is no three-line intersection)

Analysis:

First, it indicates the regular intersection of a line segment: the intersection is unique and within the two lines.

If there is a line segment UV in the input that does not conform to any other line segment, the monster must be able to go from u to v.

Therefore, we can create a graph model. If u can go to V, add an edge and BFs at last to see if it can go from the start point to the end point.

Consider the following special cases:

Although the three lines do not overlap, it does not seem to include the three-line common endpoint.

For example:

The line segments AB and BC do not conform to other line segments. Therefore, a can go to B, and B can go to C. But in fact, this triangle is closed, and the interior and exterior are not connected.

The solution is to appropriately widen the line segment (the Code allows the line segment to be extended by 1e-6 on both sides) and change it to a standard intersection.

 

Another scenario is:

When the line segments of the two endpoints are collocated, the problem also occurs. For example, it should have been closed, but it may be converted into a path because it determines whether to move forward based on the standard intersection of a line segment.

The solution is that the points on other line segments at the endpoint after the Ampersand do not participate in the diagram.

 

  1 //#define LOCAL  2 #include <cstdio>  3 #include <cstring>  4 #include <algorithm>  5 #include <iostream>  6 #include <cmath>  7 #include <vector>  8 using namespace std;  9  10 const double eps = 1e-12; 11 int dcmp(double x) 12 { 13     if(fabs(x) < eps) return 0; 14     else return x < 0 ? -1 : 1; 15 } 16  17 struct Point 18 { 19     double x, y; 20     Point(double x=0, double y=0):x(x), y(y) {} 21 }; 22 typedef Point Vector; 23  24 Point operator + (const Point& a, const Point& b) { return Point(a.x+b.x, a.y+b.y); } 25 Point operator - (const Point& a, const Point& b) { return Point(a.x-b.x, a.y-b.y); } 26 Vector operator * (const Vector& a, double p) { return Vector(a.x*p, a.y*p); } 27 Vector operator / (const Vector& a, double p) { return Vector(a.x/p, a.y/p); } 28 bool operator < (const Point& a, const Point& b) 29 { 30     return a.x < b.x || (a.x == b.x && a.y < b.y); 31 } 32 bool operator == (const Point& a, const Point& b) 33 { 34     return a.x == b.x && a.y == b.y; 35 } 36 double Dot(const Vector& a, const Vector& b) { return a.x*b.x + a.y*b.y; } 37 double Cross(const Vector& a, const Vector& b) { return a.x*b.y - a.y*b.x; } 38 double Length(const Vector& a) { return sqrt(Dot(a, a)); } 39 bool SegmentProperIntersection(const Point& a1, const Point& a2, const Point& b1, const Point& b2) 40 { 41     double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1); 42     double c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1); 43     return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3)*dcmp(c4) < 0; 44 } 45  46 bool OnSegment(const Point& p, const Point& a, const Point& b) 47 { 48     return dcmp(Cross(a-p, b-p)) == 0 && dcmp(Dot(a-p, b-p)) < 0; 49 } 50  51 const int maxn = 200 + 5; 52 int n, V; 53 int G[maxn][maxn], vis[maxn]; 54 Point p1[maxn], p2[maxn]; 55  56 bool OnAnySegment(const Point& p) 57 { 58     for(int i = 0; i < n; ++i) 59         if(OnSegment(p, p1[i], p2[i]))    return true; 60     return false; 61 } 62  63 bool IntersectionWithAnySegment(const Point& a, const Point& b) 64 { 65     for(int i = 0; i < n; ++i) 66         if(SegmentProperIntersection(a, b, p1[i], p2[i]))    return true; 67     return false; 68 } 69  70 bool dfs(int u) 71 { 72     if(u == 1)    return true; 73     vis[u] = 1; 74     for(int v = 0; v < V; ++v) 75         if(G[u][v] && !vis[v] && dfs(v))    return true; 76     return false; 77 } 78  79 bool find_path() 80 { 81     vector<Point> vertices; 82     vertices.push_back(Point(0.0, 0.0)); 83     vertices.push_back(Point(1e5, 1e5)); 84     for(int i = 0; i < n; ++i) 85     { 86         if(!OnAnySegment(p1[i])) vertices.push_back(p1[i]); 87         if(!OnAnySegment(p2[i])) vertices.push_back(p2[i]); 88     } 89     V = vertices.size(); 90     memset(G, 0, sizeof(G)); 91     memset(vis, 0, sizeof(vis)); 92     for(int i = 0; i < V; ++i) 93         for(int j = i+1; j < V; ++j) 94             if(!IntersectionWithAnySegment(vertices[i], vertices[j])) 95                 G[i][j] = G[j][i] = 1; 96     return dfs(0); 97 } 98  99 int main(void)100 {101     #ifdef LOCAL102         freopen("2797in.txt", "r", stdin);103     #endif104     105     while(scanf("%d", &n) == 1 && n)106     {107         double x1, y1, x2, y2;108         for(int i = 0; i < n; ++i)109         {110             scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);111             Point a = Point(x1, y1);112             Point b = Point(x2, y2);113             double l = Length(a-b);114             Vector v0 = (a-b) / l * 1e-6;115             p1[i] = a + v0;116             p2[i] = b - v0;117         }118         if(find_path()) puts("no");119         else puts("yes");120     }121     122     return 0;123 }

Code Jun

 

La 2797 (flat line diagram plsg) monster trap