学步园

题目：给定一个多边形A，问能否穿过多边形的洞B。

分析：计算几何、凸包、旋转卡壳、点与多边形关系。问题实际是在求多边形A的最小宽度和多边形B能容纳的最常线段长度。

1.对于多边形A，构造凸包利用，旋转卡壳求出对踵点对，然后求出最小的高。

2. 对于多边形B，所求的最长线段，一定经过多边形上的至少两个点，枚举所有点对，计算被截取的最长部分即可。

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>

using namespace std;

typedef struct pnode
{
    double x,y,d;
    pnode( double a, double b ) {x = a;y = b;}
    pnode(){};
}point;
point H[ 21 ];
point C[ 21 ];
point P0,Pn;

typedef struct lnode
{
    double x,y,dx,dy,d;
    int    id,hit,sign;
    lnode( point a, point b ) {x = a.x;y = a.y;dx = b.x-a.x;dy = b.y-a.y;}
    lnode(){};
}line;

//两点间距离 
double dist( point a, point b )
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

//点到直线距离 
double dist( point a, line l )
{
    return fabs(l.dx*(a.y-l.y)-l.dy*(a.x-l.x))/sqrt(l.dx*l.dx+l.dy*l.dy);
}

//叉乘 ab*ac 
double crossproduct( point a, point b, point c )
{
    return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);
}

//坐标排序 
bool cmp1( point a, point b )
{
    if ( a.x == b.x ) return a.y < b.y;
    else return a.x < b.x;
}

//级角排序 
bool cmp2( point a, point b )
{
    double cp = crossproduct( P0, a, b );
    if ( cp == 0 ) return a.d < b.d;
    else return cp > 0;
}

//凸包扫描算法 
double Graham( int N )
{
	sort( C+0, C+N, cmp1 );
	P0 = C[0];
    for ( int i = 1 ; i < N ; ++ i )
        C[i].d = dist( P0, C[i] );
    sort( C+1, C+N, cmp2 );
        
    //计算凸包 
    int top = 2;
    for ( int i = 3 ; i < N ; ++ i ) {
        while ( top > 0 && crossproduct( C[top-1], C[top], C[i] ) <= 0 ) -- top;
        C[++ top] = C[i];
    }
	C[++ top] = C[0];
    
    //旋转卡壳，求对踵点对 
   	int    L = 0,R = 1;
   	double D = 500.000;
   	do{
		while ( crossproduct( C[R], C[L], C[(L+1)%top] ) <= crossproduct( C[(R+1)%top], C[L], C[(L+1)%top] ) )
			R = (R+1)%top;
		
		D = min( D, dist( C[R], line( C[L], C[(L+1)%top] ) ) );
		L = (L+1)%top;
	}while ( L );
	
	return D;
}

//直线与线段相交判断 
bool l_cross_s( line b, line a )
{
    double t1 = b.dx*(a.y-b.y)-b.dy*(a.x-b.x);
    double t2 = b.dx*(a.y+a.dy-b.y)-b.dy*(a.x+a.dx-b.x);
    return t1*t2 < 0;
}

//线段相交   
bool s_cross_s( line a, line b )  
{  
	double t1 = 0.0+a.dx*(b.y-a.y)-a.dy*(b.x-a.x);  
	double t2 = 0.0+a.dx*(b.y+b.dy-a.y)-a.dy*(b.x+b.dx-a.x);  
	double t3 = 0.0+b.dx*(a.y-b.y)-b.dy*(a.x-b.x);  
	double t4 = 0.0+b.dx*(a.y+a.dy-b.y)-b.dy*(a.x+a.dx-b.x);  
	return (t1*t2 < 0)&&(t3*t4 < 0);  
}

//点在线段上   
bool on( point p, line l )  
{  
    if ( l.dx*(p.y-l.y)-l.dy*(p.x-l.x) == 0 )  
    if ( (p.x-l.x)*(p.x-l.x-l.dx) <= 0 )  
    if ( (p.y-l.y)*(p.y-l.y-l.dy) <= 0 )  
        return true;  
    return false;  
}  
  
//点在多边形内  
bool in( point p, point* P, int n )  
{  
    double d[4][2] = {-101,-103,-103,101,101,-103,101,103};  
    for ( int t = 0 ; t < 4 ; ++ t ) {  
        line s1 = line( p, point( d[t][0], d[t][1] ) );  
        int  count = 0;  
        for ( int i = 0 ; i < n ; ++ i ) {  
            line s2 = line( P[i], P[i+1] );  
            if ( on( p, s2 ) ) return true;  
            if ( s_cross_s( s1, s2 ) ) count ++;  
            if ( on( P[i], s1 ) && l_cross_s( s1, line( P[i+1], P[(i-1+n)%n] ) ) ) count ++;  
        }
        if ( count%2 == 0 ) return false;   
    }  
    return true;  
}  

//两直线交点 
point crosspoint( line l, line m )
{
    point a = point( m.x, m.y );
    point b = point( m.x+m.dx, m.y+m.dy );
    if ( m.dx*l.dy == m.dy*l.dx ) {
        if ( dist( point( l.x, l.y ), a ) < dist( point( l.x, l.y ), b ) )
            return a;
        else return b;
    }else {
        double a1 = -l.dy,b1 = l.dx,c1 = l.dx*l.y-l.dy*l.x;
        double a2 = -m.dy,b2 = m.dx,c2 = m.dx*m.y-m.dy*m.x;
        double x = (c1*b2-c2*b1)/(a1*b2-a2*b1);
        double y = (c1*a2-c2*a1)/(b1*a2-b2*a1);
        return point( x, y );
    } 
}

//计算空的最大长度 
double Calcul( int N )
{
	H[N] = H[0];
	
	point  X[ 21 ];
	double D = 0.0;
	for ( int i = 0 ; i < N ; ++ i )
	for ( int j = i+1 ; j < N ; ++ j ) {
		line l = line( H[i], H[j] );
			
		int S = 0;
		for ( int k = 0 ; k < N ; ++ k ) {
			line m = line( H[k], H[k+1] );
			if ( l_cross_s( l, m ) )
				X[S ++] = crosspoint( l, m );
		}
		X[S ++] = H[i];
		X[S ++] = H[j];
		
		P0 = point( -101, -103 );
		for ( int k = 0 ; k < S ; ++ k )
			X[k].d = dist( P0, X[k] );
		sort( X, X+S, cmp2 );
		
		double sum = 0.0;
		int    fla = 0;
		for ( int i = 1 ; i < S ; ++ i ) {
			if ( in( point( (X[i-1].x+X[i].x)/2, (X[i-1].y+X[i].y)/2 ), H, N ) ) {
				if ( fla ) sum += dist( X[i-1], X[i] );
				else sum = dist( X[i-1], X[i] );
				D = max( D, sum );
				fla = 1;
			}
		}
	}
	
	return D;
}

int main()
{
    int T,N,M;
    while ( scanf("%d",&T) != EOF ) 
    while ( T -- ) {
        scanf("%d",&N);
        for ( int i = 0 ; i < N ; ++ i )
            scanf("%lf%lf",&H[i].x,&H[i].y);
        scanf("%d",&M);
        for ( int i = 0 ; i < M ; ++ i )
            scanf("%lf%lf",&C[i].x,&C[i].y);
        
        //计算硬币最小宽度 
		double d = Graham( M );
        
        //计算空的最大长度 
        double D = Calcul( N );
        
        if ( d <= D ) printf("legal\n");
        else printf("illegal\n");
    }
    return 0;
}