4.1 求解平面最远点对（POJ 2187 Beauty Contest）
struct Point
{
int x,y;
Point(int _x = 0,int _y = 0)
{
x = _x; y = _y;
}
Point operator -(const Point &b)const
{
return Point(x - b.x, y - b.y);
}
int operator ^(const Point &b)const
{
return x*b.y - y*b.x;
}
int operator *(const Point &b)const
{
return x*b.x + y*b.y;
}
void input()
{
scanf("%d%d",&x,&y);
}
};
//距离的平方
int dist2(Point a,Point b)
{
return (a-b)*(a-b);
}
//******二维凸包，int***********
const int MAXN = 50010;
Point list[MAXN];
int Stack[MAXN],top;
bool _cmp(Point p1,Point p2)
{
int tmp = (p1-list[0])^(p2-list[0]);
if(tmp > 0)return true;
else if(tmp == 0 && dist2(p1,list[0]) <= dist2(p2,list[0]))
return true;
else return false;
}
void Graham(int n)
{
Point p0;
int k = 0;
p0 = list[0];
for(int i = 1;i < n;i++)
if(p0.y > list[i].y || (p0.y == list[i].y && p0.x > list[i].x))
{
p0 = list[i];
k = i;
}
swap(list[k],list[0]);
sort(list+1,list+n,_cmp);
if(n == 1)
{
top = 1;
Stack[0] = 0;
return;
}
if(n == 2)
{
top = 2;
Stack[0] = 0; Stack[1] = 1;
return;
}
Stack[0] = 0; Stack[1] = 1;
top = 2;
for(int i = 2;i < n;i++)
{
while(top > 1 &&
((list[Stack[top-1]]-list[Stack[top-2]])^(list[i]-list[Stack[top-2]])) <= 0)
top--;
Stack[top++] = i;
}
}
//旋转卡壳，求两点间距离平方的最大值
int rotating_calipers(Point p[],int n)
{
int ans = 0;
Point v;
int cur = 1;
for(int i = 0;i < n;i++)
{
v = p[i]-p[(i+1)%n];
while((v^(p[(cur+1)%n]-p[cur])) < 0)
cur = (cur+1)%n;
ans = max(ans,max(dist2(p[i],p[cur]),dist2(p[(i+1)%n],p[(cur+1)%n])));
}
return ans;
}
Point p[MAXN];
int main()
{
int n;
while(scanf("%d",&n) == 1)
{
for(int i = 0;i < n;i++)list[i].input();
Graham(n);
for(int i = 0;i < top;i++)p[i] = list[Stack[i]];
printf("%d
",rotating_calipers(p,top));
}
return 0;
}
4.2 求解平面点集最大三角形
//旋转卡壳计算平面点集最大三角形面积
int rotating_calipers(Point p[],int n)
{
int ans = 0;
Point v;
for(int i = 0;i < n;i++)
{
int j = (i+1)%n;
int k = (j+1)%n;
while(j != i && k != i)
{
ans = max(ans,abs((p[j]-p[i])^(p[k]-p[i])));
while( ((p[i]-p[j])^(p[(k+1)%n]-p[k])) < 0 )
k = (k+1)%n;
j = (j+1)%n;
}
}
return ans;
}
Point p[MAXN];
int main()
{
int n;
while(scanf("%d",&n) == 1)
{
if(n == -1)break;
for(int i = 0;i < n;i++)list[i].input();
Graham(n);
for(int i = 0;i < top;i++)p[i] = list[Stack[i]];
printf("%.2f
",(double)rotating_calipers(p,top)/2);
}
return 0;
}
4.3 求解两凸包最小距离（POJ 3608）
const double eps = 1e-8;
int sgn(double x)
{
if(fabs(x) < eps)return 0;
if(x < 0)return -1;
else return 1;
}
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);
}
double operator ^(const Point &b)const
{
return x*b.y - y*b.x;
}
double operator *(const Point &b)const
{
return x*b.x + y*b.y;
}
void input()
{
scanf("%lf%lf",&x,&y);
}
};
struct Line
{
Point s,e;
Line(){}
Line(Point _s,Point _e)
{
s = _s; e = _e;
}
};
//两点间距离
double dist(Point a,Point b)
{
return sqrt((a-b)*(a-b));
}
//点到线段的距离，返回点到线段最近的点
Point NearestPointToLineSeg(Point P,Line L)
{
Point result;
double t = ((P-L.s)*(L.e-L.s))/((L.e-L.s)*(L.e-L.s));
if(t >=0 && t <= 1)
{
result.x = L.s.x + (L.e.x - L.s.x)*t;
result.y = L.s.y + (L.e.y - L.s.y)*t;
}
else
{
if(dist(P,L.s) < dist(P,L.e))
result = L.s;
else result = L.e;
}
return result;
}
/*
* 求凸包，Graham算法
* 点的编号0~n-1
* 返回凸包结果Stack[0~top-1]为凸包的编号
*/
const int MAXN = 10010;
Point list[MAXN];
int Stack[MAXN],top;
//相对于list[0]的极角排序
bool _cmp(Point p1,Point p2)
{
double tmp = (p1-list[0])^(p2-list[0]);
if(sgn(tmp) > 0)return true;
else if(sgn(tmp) == 0 && sgn(dist(p1,list[0]) - dist(p2,list[0])) <= 0)
return true;
else return false;
}
void Graham(int n)
{
Point p0;
int k = 0;
p0 = list[0];
//找最下边的一个点
for(int i = 1;i < n;i++)
{
if( (p0.y > list[i].y) || (p0.y == list[i].y && p0.x > list[i].x) )
{
p0 = list[i];
k = i;
}
}
swap(list[k],list[0]);
sort(list+1,list+n,_cmp);
if(n == 1)
{
top = 1;
Stack[0] = 0;
return;
}
if(n == 2)
{
top = 2;
Stack[0] = 0;
Stack[1] = 1;
return ;
}
Stack[0] = 0;
Stack[1] = 1;
top = 2;
for(int i = 2;i < n;i++)
{
while(top > 1 &&
sgn((list[Stack[top-1]]-list[Stack[top-2]])^(list[i]-list[Stack[top-2]])) <=
0)
top--;
Stack[top++] = i;
}
}
//点p0到线段p1p2的距离
double pointtoseg(Point p0,Point p1,Point p2)
{
return dist(p0,NearestPointToLineSeg(p0,Line(p1,p2)));
}
//平行线段p0p1和p2p3的距离
double dispallseg(Point p0,Point p1,Point p2,Point p3)
{
double ans1 = min(pointtoseg(p0,p2,p3),pointtoseg(p1,p2,p3));
double ans2 = min(pointtoseg(p2,p0,p1),pointtoseg(p3,p0,p1));
return min(ans1,ans2);
}
//得到向量a1a2和b1b2的位置关系
double Get_angle(Point a1,Point a2,Point b1,Point b2)
{
return (a2-a1)^(b1-b2);
}
double rotating_calipers(Point p[],int np,Point q[],int nq)
{
int sp = 0, sq = 0;
for(int i = 0;i < np;i++)
if(sgn(p[i].y - p[sp].y) < 0)
sp = i;
for(int i = 0;i < nq;i++)
if(sgn(q[i].y - q[sq].y) > 0)
sq = i;
double tmp;
double ans = dist(p[sp],q[sq]);
for(int i = 0;i < np;i++)
{
while(sgn(tmp = Get_angle(p[sp],p[(sp+1)%np],q[sq],q[(sq+1)%nq])) < 0)
sq = (sq+1)%nq;
if(sgn(tmp) == 0)
ans = min(ans,dispallseg(p[sp],p[(sp+1)%np],q[sq],q[(sq+1)%nq]));
else ans = min(ans,pointtoseg(q[sq],p[sp],p[(sp+1)%np]));
sp = (sp+1)%np;
}
return ans;
}
double solve(Point p[],int n,Point q[],int m)
{
return min(rotating_calipers(p,n,q,m),rotating_calipers(q,m,p,n));
}
Point p[MAXN],q[MAXN];
int main()
{
int n,m;
while(scanf("%d%d",&n,&m) == 2)
{
if(n == 0 && m == 0)break;
for(int i = 0;i < n;i++)
list[i].input();
Graham(n);
for(int i = 0;i < top;i++)
p[i] = list[i];
n = top;
for(int i = 0;i < m;i++)
list[i].input();
Graham(m);
for(int i = 0;i < top;i++)
q[i] = list[i];
m = top;
printf("%.4f
",solve(p,n,q,m));
}
return 0;
}