poj 3608 Bridge Across Islands

题目：计算两个不相交凸多边形间的最小距离。

分析：计算几何、凸包、旋转卡壳。分别求出凸包，利用旋转卡壳求出对踵点对，枚举距离即可。

注意：1.利用向量法判断旋转，而不是计算角度；避免精度问题和TLE。

            2.遇到平行线段时，需要计算4组点到线段距离，不然会漏掉对踵点对。

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

using namespace std;

//点结构 
typedef struct pnode
{
	double x,y,d;
	pnode( double a, double b ) {x = a;y = b;}  
    pnode(){}; 
}point;
point T,P[10005],Q[10005];

//线段结构 
typedef struct lnode  
{  
    double x,y,dx,dy;
    lnode( point a, point b ) {x = a.x;y = a.y;dx = b.x-a.x;dy = b.y-a.y;} 
    lnode(){};  
}line;  

//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);
}

//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));
}

//点到线段距离 
double dist( point a, point p, point q )  
{
    line l = line( p, q );
    //判断垂足位置 
    if ( (l.dx*(p.x-a.x)+l.dy*(p.y-a.y))*(l.dx*(q.x-a.x)+l.dy*(q.y-a.y)) < 0 )
    	return fabs(l.dx*(a.y-l.y)-l.dy*(a.x-l.x))/sqrt(l.dx*l.dx+l.dy*l.dy); 
	else return min( dist( a, p ), dist( a, q ) );
}

//坐标比较 
bool cmp1( point a, point b )
{
	return (a.x==b.x)?(a.y<b.y):(a.x<b.x);
}

//级角比较 
bool cmp2( point a, point b )
{
	double cp = crossproduct( T, a, b );
	if ( !cp ) return a.d < b.d;
	else return cp > 0;
}

//凸包 
int graham( point* p, int n )
{
	sort( p+0, p+n, cmp1 );
	for ( int i = 1 ; i < n ; ++ i )
		p[i].d = dist( p[0], p[i] );
	T = p[0];
	sort( p+1, p+n, cmp2 );
	
	int top = 1;
	for ( int i = 2 ; i < n ; ++ i ) {
		while ( top > 0 && crossproduct( p[top-1], p[top], p[i] ) <= 0 ) -- top;
		p[++ top] = p[i];
	}
	p[++ top] = p[0];
	
	return top;
}

//利用向量判断夹角 
double judge( point a, point b, point c, point d )
{
	return crossproduct( c, d, point( c.x+b.x-a.x, c.y+b.y-a.y ) );
}

//旋转卡壳 
double rotatingcalipers( point* p, point* q, int n, int m )
{
	double D = 30000.0;
	int R = 0;
	for ( int i = 0 ; i < m ; ++ i )
		if ( q[i].x >= q[R].x ) R = i;
	for ( int L = 0 ; L < n ; ++ L ) {
		while ( judge( p[L], p[L+1], q[R], q[R+1] ) < 1e-6 )
			R = (R+1)%m;
		//两条边平行时，需计算平行线段间最短距离，即四个点到线段的距离 
		D = min( min( D, dist( p[L], q[R] ) ), 
				 min( min( dist( p[L], q[R], q[R+1] ), dist( q[R], p[L], p[L+1] ) ),
				 	  min( dist( p[L+1], q[R], q[R+1] ), dist( q[R+1], p[L], p[L+1] ) ) ) );
	}
	
	return D;
}

int main()
{
	int N,M;
	while ( scanf("%d%d",&N,&M) && N ) {
		for ( int i = 0 ; i < N ; ++ i )
			scanf("%lf%lf",&P[i].x,&P[i].y);
		for ( int i = 0 ; i < M ; ++ i )
			scanf("%lf%lf",&Q[i].x,&Q[i].y);
			
		N = graham( P, N );
		M = graham( Q, M );
		
		printf("%.5lf\n",rotatingcalipers( P, Q, N, M ));
	}
	return 0;
}