URAL 1990 Podracing

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>

using namespace std;

#define LL long long
#define eps 1e-12
#define mnx 100020
#define mxe 2000020
#define inf 1e12
#define Pi acos( -1.0 )

//精度
int dcmp( double x ){
    if( fabs( x ) < eps ) return 0;
    return x < 0 ? -1 : 1;
}
// 点
struct point{
    double x, y;
    point( double x = 0, double y = 0 ) : x(x), y(y) {}
    point operator + ( const point &b ) const{
        return point( x + b.x, y + b.y );
    }
    point operator - ( const point &b ) const{
        return point( x - b.x, y - b.y );
    }
    point operator * ( const double &k ) const{
        return point( x * k, y * k );
    }
    point operator / ( const double &k ) const{
        return point( x / k, y / k );
    }
    bool operator < ( const point &b ) const{
        return dcmp( y - b.y ) < 0 || dcmp( y - b.y ) == 0 && dcmp( x - b.x ) < 0;
    }
    bool operator == ( const point &b ) const{
        return dcmp( x - b.x ) == 0 && dcmp( y - b.y ) == 0;
    }
    double len(){
        return sqrt( x * x + y * y );
    }
    void input(){
        scanf( "%lf%lf", &x, &y );
    }
};
typedef point Vector;
// 点积
double dot( Vector a, Vector b ){
    return a.x * b.x + a.y * b.y;
}
// 叉积
double cross( Vector a, Vector b ){
    return a.x * b.y - a.y * b.x;
}
int n, m, q;
point a[mnx], b[mnx], c[mnx];
double calc( point p0, point p1, point p2  ){
    if(dcmp(p1.y - p2.y) == 0) {
        return min((p0 - p1).len(), (p0 - p2).len());
    }
    if( p1.y > p2.y ) swap( p1, p2 );
    double t = (p0.y - p1.y) / ( p2.y - p1.y );
    point v = p1 + ( p2 - p1 ) * t;
    return ( p0 - v ).len();
}
void solve(){
    double ans = inf;
    int j = 0;
    for( int i = 0; i < n; i++ ){
        while( j < m-1 ){
            if( dcmp(a[i].y-b[j].y) >= 0 && dcmp(a[i].y-b[j+1].y) <= 0 ){
                ans = min( ans, calc( a[i], b[j], b[j+1] ) );
                break;
            }
            else j++;
        }
    }
    j = 0;
    for( int i = 0; i < m; i++ ){
        while( j < n-1 ){
            if( dcmp(b[i].y-a[j].y) >= 0 && dcmp(b[i].y-a[j+1].y) <= 0 ){
                ans = min( ans, calc( b[i], a[j], a[j+1] ) );
                break;
            }
            else j++;
        }
    }
    sort( c, c + q );
    int cnt = 0;
    for( int i = 0; i < q; i++ ){
        if( dcmp(c[i].y-a[0].y) < 0 || dcmp(c[i].y-a[n-1].y) > 0 ) continue;
        c[cnt++] = c[i];
    }
    q = cnt;
    int i = 0; j = 0;
    for( int k = 0; k < q; k++ ){
        double res = -inf;
        int cur = k;
        while( cur < q && dcmp(c[cur+1].y - c[cur].y) == 0 ) cur++;
        int tmp = cur;

        while( i < n-1 ){
            if( dcmp(c[k].y-a[i].y) >= 0 && dcmp(c[k].y-a[i+1].y) <= 0 )
                break;
            else i++;
        }
        while( j < m-1 ){
            if( dcmp(c[k].y-b[j].y) >= 0 && dcmp(c[k].y-b[j+1].y) <= 0 )
                break;
            else j++;
        }//cout << k << " " << cur << endl;
        for( int u = k; u < cur; u++ ){
            if( cross( c[u] - a[i], a[i+1] - a[i] ) < 0 ) k++;
            else break;
        }
        for( int u = cur; u > k; u-- ){
            if( cross( c[u] - b[j], b[j+1] - b[j] ) > 0 ) cur--;
            else break;
        }
        res = max( res, calc( c[k], a[i], a[i+1] ) );
        res = max( res, calc( c[cur], b[j], b[j+1] ) );
        //printf( "%.5lf\n", res );
        for( int u = k; u < cur; u++ ){
            res = max( res, ( c[u+1] - c[u] ).len() );
        }
        ans = min( res, ans );
        k = tmp;
    }
    printf( "%.8lf\n", ans );
}
int main(){
    while( scanf( "%d", &n ) != EOF ){
        for( int i = 0; i < n; i++ )
            a[i].input();
        scanf( "%d", &m );
        for( int i = 0; i < m; i++ )
            b[i].input();
        scanf( "%d", &q );
        for( int i = 0; i < q; i++ )
            c[i].input();
        solve();
    }
    return 0;
}