BZOJ1074 [SCOI2007]折纸origami

原文地址http://www.cnblogs.com/Christopher-Cao/p/5482575.html

这道题是计算几何的裸题。看起来折叠了几次要用什么东西来维护，实际上因为n<=8所以可以暴力统计。唯一有难度的东西是找出一个点的对称点（实际上也没有什么难度吧）

贴代码

#include<cstdio>
#include<cmath>
#include<vector>
using namespace std ; 

const int MAXN = 10 ; 
const int MAXM = 60 ; 
const double eps = 1e-6 ; 
int N , M ;

struct Point {
    double x , y ; 
    Point ( const double x , const double y ) :
    x ( x ) , y ( y ) {} ;
} ; 

struct Vector {
    double x , y ; 
    Vector ( const double x , const double y ) : 
    x ( x ) , y ( y ) {} ; 
} ; 

struct Line {
    Point From , To ; 
    Line ( const double x1 , const double y1 , 
           const double x2 , const double y2 ) :
    From ( x1 , y1 ) , To ( x2 , y2 ) {} ;
    Vector toVector () const ;
} ;

Vector operator - ( const Point & First , const Point & Second ) {
    return Vector ( First . x - Second . x , First . y - Second . y ) ; 
} 

Point operator - ( const Point & First , const Vector & Second ) {
    return Point ( First . x - Second . x , First . y - Second . y ) ; 
} 

Vector operator * ( const double k , const Vector & Input ) {
    return Vector ( k * Input . x , k * Input . y ) ; 
}

Point operator + ( const Point & First , const Vector & Second ) {
    return Point ( First . x + Second . x , First . y + Second . y ) ; 
}

//上面是向量的定义和各种加减运算
double operator ^ ( const Vector & First , const Vector & Second ) {
    return First . x * Second . y - First . y * Second . x ; 
}  
//叉积
double operator * ( const Vector & First , const Vector & Second ) {
    return First . x * Second . x + First . y * Second . y ; 
}
//内积
double abs ( const Vector & Input ) {
    return sqrt ( Input . x * Input . x + Input . y * Input . y ) ; 
}
//模
Vector Line :: toVector () const { return To - From ; }  

vector < Line > Opts ; 

Point reflect ( const Point o , const Line L ) {
    const double dis = ( ( L . From - o ) ^ ( L . To - o ) ) / abs ( L . toVector () ) ; 
    const Vector Lp = 1.0 / abs ( L . toVector () ) * L . toVector () ; 
    return o + 2.0 * dis * Vector ( Lp . y , - Lp . x ) ;
}

int Query ( const Point o , const int T ) {
    //printf ( "Dfs %.2lf %.2lf %d\n" , o . x , o . y , T ) ;
    if ( T == -1 ) 
        return ( eps <= o . x && o . x <= 100.0 - eps ) && 
               ( eps <= o . y && o . y <= 100.0 - eps ) ; 
    //这个地方一开始的时候没有加eps所以wa了（良心样例）
    const double CrossValue = 
        ( ( Opts [ T ] . From - o ) ^ ( Opts [ T ] . To - o ) ) ;
    if ( CrossValue < - eps ) return 0 ;     
    if ( - eps <= CrossValue && CrossValue <= eps ) return 0 ; 
    //这个地方多了一行是因为一开始读题的时候以为边界上只算一次，后来没过样例才明白边界不算
    return Query ( o , T - 1 ) + 
           Query ( reflect ( o , Opts [ T ] ) , T - 1 ) ; 
}

int main () {
    scanf ( "%d" , & N ) ; 
    for ( int i = 0 ; i < N ; ++ i ) {
        double x1 , y1 , x2 , y2 ; 
        scanf ( "%lf%lf%lf%lf" , & x1 , & y1 , & x2 , & y2 ) ; 
        Opts . push_back ( Line ( x1 , y1 , x2 , y2 ) ) ; 
    } 
    scanf ( "%d" , & M ) ; 
    while ( M -- ) {
        double x , y ; 
        scanf ( "%lf%lf" , & x , & y ) ; 
        printf ( "%d\n" , Query ( Point ( x , y ) , N - 1 ) ) ;
    }
    return 0 ;
}