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。
[cpp]
#include
#include
#include
#include
#include
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;