题目：http://www.lydsy.com/JudgeOnline/problem.php?id=2433
我们可以YY出必定存在最短路仅由S,T,和公共点之间的直线连边构成，那么就每次对于一个点，求出其到右边所有点之间的可行连边，这个可以维护两个斜率上下界，为了保证精度，使用向量来表示，然后要是S在右边，那就和T交换一下，最后最短路一次即可,O(n^2 + n^2 log n)
代码（计算几何太弱了，调了整整两个小时QaQ）：
#include
#include
#include
#include
#include
#include
using namespace std ;
#define travel( x ) for ( vector < edge > :: iterator p = E[ x ].begin( ) ; p != E[ x ].end( ) ; ++ p )
#define rep( i , x ) for ( int i = 0 ; i ++ < x ; )
#define REP( i , l , r ) for ( int i = l ; i <= r ; ++ i )
#define addedge( s , t , d ) E[ s ].push_back( edge( t , d ) )
const int maxn = 2010 , maxv = 4010 , inf = 0x7fffffff ;
const double esp = 0.0000000000001 ;
inline int cmp( double x , double y ) {
return y - x > esp ? -1 : ( x - y > esp ) ;
}
struct edge {
int t ; double d ;
edge( int _t , double _d ) : t( _t ) , d( _d ) {
}
} ;
vector < edge > E[ maxv ] ;
double dist[ maxv ] ;
bool f[ maxv ] ;
int S , T , V = 0 , node[ maxn ][ 2 ] ;
struct Cmp {
bool operator ( ) ( int x , int y ) {
return dist[ x ] > dist[ y ] ;
}
} ;
priority_queue < int , vector < int > , Cmp > q ;
inline double dijkstra( ) {
rep( i , V ) dist[ i ] = inf , f[ i ] = true ;
dist[ S ] = 0 , q.push( S ) ;
int now ; double cost ;
while ( ! q.empty( ) ) {
now = q.top( ) ; q.pop( ) ;
if ( ! f[ now ] ) continue ;
f[ now ] = false ;
if ( now == T ) return dist[ T ] ;
travel( now ) if ( cmp( dist[ p -> t ] , cost = dist[ now ] + p -> d ) == 1 ) {
dist[ p -> t ] = cost , f[ p -> t ] = true , q.push( p -> t ) ;
}
}
return dist[ T ] ;
}
struct Point {
double x , y ;
void Read( ) {
scanf( "%lf%lf" , &x , &y ) ;
}
} ;
struct Vector {
double x , y ;
Vector( double _x , double _y ) : x( _x ) , y( _y ) {
}
} ;
Vector operator - ( const Point &a , const Point &b ) {
return Vector( a.x - b.x , a.y - b.y ) ;
}
double operator * ( const Vector &a , const Vector &b ) {
return a.x * b.y - b.x * a.y ;
}
inline double sqr( double val ) {
return val * val ;
}
inline double Dist( const Point &a , const Point &b ) {
return sqrt( sqr( a.x - b.x ) + sqr( a.y - b.y ) ) ;
}
Point st , to , pos[ maxn ][ 2 ] , px[ maxn ][ 2 ] ;
double sp , ans ;
int n , ps = 0 , pt = 0 ;
inline void make_edge( Point P , int N , int M ) {
Vector L = Vector( 0 , 1 ) , R = Vector( 0 , -1 ) ;
REP( i , N , pt ) {
if ( cmp( ( px[ i ][ 0 ] - P ) * L , 0 ) != -1 && cmp( R * ( px[ i ][ 0 ] - P ) , 0 ) != -1 ) {
addedge( M , node[ i ][ 0 ] , Dist( P , px[ i ][ 0 ] ) ) ;
L = ( px[ i ][ 0 ] - P ) ;
} else if ( cmp( ( px[ i ][ 0 ] - P ) * R , 0 ) == 1 ) break ;
if ( cmp( ( px[ i ][ 1 ] - P ) * L , 0 ) != -1 && cmp( R * ( px[ i ][ 1 ] - P ) , 0 ) != -1 ) {
addedge( M , node[ i ][ 1 ] , Dist( P , px[ i ][ 1 ] ) ) ;
R = ( px[ i ][ 1 ] - P ) ;
} else if ( cmp( L * ( px[ i ][ 1 ] - P ) , 0 ) == 1 ) break ;
}
}
int main( ) {
scanf( "%d" , &n ) ;
rep( i , n ) pos[ i ][ 0 ].Read( ) , pos[ i ][ 1 ].Read( ) ;
st.Read( ) ; to.Read( ) ; scanf( "%lf" , &sp ) ;
if ( st.x > to.x ) swap( st , to ) ;
rep( i , n ) {
if ( ! ps ) {
if ( cmp( st.x , pos[ i ][ 0 ].x ) != -1 && cmp( st.x , pos[ i ][ 1 ].x ) != 1 && cmp( st.y , pos[ i ][ 0 ].y ) != -1 && cmp( st.y , pos[ i ][ 1 ].y ) != 1 ) {
ps = i ;
}
}
if ( ! pt ) {
if ( cmp( to.x , pos[ i ][ 0 ].x ) != -1 && cmp( to.x , pos[ i ][ 1 ].x ) != 1 && cmp( to.y , pos[ i ][ 0 ].y ) != -1 && cmp( to.y , pos[ i ][ 1 ].y ) != 1 ) {
pt = i ;
}
}
}
if ( ps == pt ) ans = Dist( st , to ) ; else {
REP( i , ps , ( pt - 1 ) ) {
px[ i ][ 0 ].x = px[ i ][ 1 ].x = pos[ i ][ 1 ].x ;
px[ i ][ 0 ].y = min( pos[ i ][ 1 ].y , pos[ i + 1 ][ 1 ].y ) ;
px[ i ][ 1 ].y = max( pos[ i ][ 0 ].y , pos[ i + 1 ][ 0 ].y ) ;
}
px[ pt ][ 0 ] = px[ pt ][ 1 ] = to ;
REP( i , ps , ( pt - 1 ) ) {
node[ i ][ 0 ] = ++ V , node[ i ][ 1 ] = ++ V ;
}
S = ++ V , node[ pt ][ 0 ] = node[ pt ][ 1 ] = T = ++ V ;
make_edge( st , ps , S ) ;
REP( i , ps , ( pt - 1 ) ) {
make_edge( px[ i ][ 0 ] , i , node[ i ][ 0 ] ) , make_edge( px[ i ][ 1 ] , i , node[ i ][ 1 ] ) ;
}
ans = dijkstra( ) ;
}
printf( "%.10f\n" , ans / sp ) ;
return 0 ;
}