POJ (线段相交 最短路) The Doors

题意：

一个正方形中有n道竖直的墙，每道墙上开两个门。求从左边中点走到右边中点的最短距离。

分析：

以起点终点和每个门的两个端点建图，如果两个点可以直接相连(即不会被墙挡住)，则权值为两点间的欧几里得距离。

然后求起点到终点的最短路即可。

 #include <cstdio>
 #include <cmath>
 #include <cstring>
 #include <vector>
 #include <algorithm>
 using namespace std;
 
 const int maxn = ;
 const double INF = 1e4;
 const double eps = 1e-;
 
 struct Point
 {
     double x, y;
     Point(double x=, double y=):x(x), y(y) {}
 }p[maxn * ];
 
 typedef Point Vector;
 
 Point read_point()
 {
     double x, y;
     scanf("%lf%lf", &x, &y);
     return Point(x, y);
 }
 
 Point operator - (const Point& A, const Point& B)
 { return Point(A.x-B.x, A.y-B.y); }
 
 Vector operator / (const Vector& A, double p)
 { return Vector(A.x/p, A.y/p); }
 
 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)); }
 
 struct Door
 {
     double x, y1, y2, y3, y4;
     Door(double x=, double y1=, double y2=, double y3=, double y4=):x(x), y1(y1), y2(y2), y3(y3), y4(y4) {}
 };
 
 vector<Door> door;
 
 double d[maxn * ], w[maxn * ][maxn * ];
 bool vis[maxn * ];
 int cnt;
 
 bool isOK(int a, int b)
 {//判断两点是否能直接到达
     if(p[a].x >= p[b].x) swap(a, b);
     for(int i = ; i < door.size(); ++i)
     {
         if(door[i].x <= p[a].x) continue;
         if(door[i].x >= p[b].x) break;
         double k = (p[b].y-p[a].y) / (p[b].x-p[a].x);
         double y = p[a].y + k * (door[i].x - p[a].x);
         if(!(y>=door[i].y1&&y<=door[i].y2 || y>=door[i].y3&&y<=door[i].y4)) return false;
     }
     return true;
 }
 
 void Init()
 {
     for(int i = ; i < cnt; ++i)
         for(int j = i; j < cnt; ++j)
             if(i == j) w[i][j] = ;
             else w[i][j] = w[j][i] = INF;
 }
 
 int main()
 {
     //freopen("in.txt", "r", stdin);
 
     int n;
     while(scanf("%d", &n) ==  && n + )
     {
         door.clear();
         memset(vis, false, sizeof(vis));
         memset(d, , sizeof(d));
 
         p[] = Point(, );
         cnt = ;
         for(int i = ; i < n; ++i)
         {
             double x, y[];
             scanf("%lf", &x);
             for(int j = ; j < ; ++j) { scanf("%lf", &y[j]); p[cnt++] = Point(x, y[j]); }
             door.push_back(Door(x, y[], y[], y[], y[]));
         }
         p[cnt++] = Point(, );
 
         Init();
 
         for(int i = ; i < cnt; ++i)
             for(int j = i+; j < cnt; ++j)
             {
                 double l = Length(Vector(p[i]-p[j]));
                 if(p[i].x == p[j].x) continue;
                 if(isOK(i, j))
                     w[i][j] = w[j][i] = l;
             }
         //Dijkstra
         d[] = ;
         for(int i = ; i < cnt; ++i) d[i] = INF;
         for(int i = ; i < cnt; ++i)
         {
             int x;
             double m = INF;
             for(int y = ; y < cnt; ++y) if(!vis[y] && d[y] <= m) m = d[x=y];
             vis[x] = ;
             for(int y = ; y < cnt; ++y) d[y] = min(d[y], d[x] + w[x][y]);
         }
 
         printf("%.2f\n", d[cnt-]);
     }
 
     return ;
 }

代码君