题目：http://www.lydsy.com/JudgeOnline/problem.php?id=1027
额。。。计算几何太弱了，这题搞了N久才A掉，就是用最小环求一下最小的凸包，然后记得要特判一下所有点都在一个点处的情况。。。
代码：
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <cmath>
using namespace std ;
const int inf = 1000000 ;
const double esp = 0.00000000001 ;
const int maxn = 510 ;
struct Point {
double x , y ;
void read( ) {
scanf( "%lf%lf" , &x , &y ) ;
}
} p0[ maxn ] , p1[ maxn ] ;
struct Vector {
double x , y ;
} ;
Vector operator - ( const Point &x , const Point &y ) {
return ( Vector ) { x.x - y.x , x.y - y.y } ;
}
double operator * ( const Vector &x , const Vector &y ) {
return x.x * y.y - y.x * x.y ;
}
double mul( const Vector &x , const Vector &y ) {
return x.x * y.x + x.y * y.y ;
}
struct graph {
vector < int > E[ maxn ] ;
int V ;
void Init( int _V ) {
V = _V ;
}
void addedge( int s , int t ) {
E[ s ].push_back( t ) ;
}
int dist[ maxn ] ;
queue < int > q ;
void bfs( int v ) {
for ( int i = 0 ; i ++ < V ; ) dist[ i ] = inf ;
for ( vector < int > :: iterator p = E[ v ].begin( ) ; p != E[ v ].end( ) ; ++ p ) {
dist[ *p ] = 1 , q.push( *p ) ;
}
while ( ! q.empty( ) ) {
int now = q.front( ) ; q.pop( ) ;
for ( vector < int > :: iterator p = E[ now ].begin( ) ; p != E[ now ].end( ) ; ++ p ) if ( dist[ now ] + 1 < dist[ *p ] ) {
dist[ *p ] = dist[ now ] + 1 ;
q.push( *p ) ;
}
}
}
} g ;
int n , m ;
bool cmp( double x , double y ) {
return abs( x - y ) < esp ;
}
int main( ) {
scanf( "%d%d" , &n , &m ) ;
for ( int i = 0 ; i ++ < n ; ) {
p0[ i ].read( ) ;
double x ; scanf( "%lf" , &x ) ;
}
bool flag = true ;
for ( int i = 0 ; i ++ < m ; ) {
p1[ i ].read( ) ;
double x ; scanf( "%lf" , &x ) ;
if ( ! ( cmp( p1[ i ].x , p1[ 1 ].x ) && cmp( p1[ i ].y , p1[ 1 ].y ) ) ) flag = false ;
}
if ( flag ) {
for ( int i = 0 ; i ++ < n ; ) if ( cmp( p0[ i ].x , p1[ 1 ].x ) && cmp( p0[ i ].y , p1[ 1 ].y ) ) {
printf( "1\n" ) ;
return 0 ;
}
}
g.Init( n ) ;
for ( int i = 0 ; i ++ < n ; ) for ( int j = 0 ; j ++ < n ; ) if ( i != j ) {
Vector v = p0[ j ] - p0[ i ] ;
flag = true ;
for ( int k = 0 ; k ++ < m ; ){
double ret = v * ( p1[ k ] - p0[ i ] ) ;
if ( ! ( ret > esp || ( ret > - esp && ret < esp && mul( p1[ k ] - p0[ i ] , p1[ k ] - p0[ j ] ) < esp ) ) ) {
flag = false ;
break ;
}
}
if ( flag ) g.addedge( i , j ) ;
}
int ans = inf ;
for ( int i = 0 ; i ++ < n ; ) {
g.bfs( i ) ;
ans = min( ans , g.dist[ i ] ) ;
}
printf( "%d\n" , ans < inf ? ans : -1 ) ;
return 0 ;
}
