UVA 10173 最小矩形覆盖（凸包+旋转卡壳）

传送门

题意正如标题。

可为什么用旋转卡壳呢？

先把有的点用一个凸包维护，那么这个凸包一定覆盖了所有的点，如果要有一个矩形覆盖所有的点，

那么他一定有一条边和凸包的一条边重合。

#include<bits/stdc++.h>
#define eps 1e-8
using namespace std;

const int _max = 1e3 + 10;
const double PI = acos(-1);
int n;

int sgn(double x) //三态函数
{
if(fabs(x)<eps) return 0;
else return x<0?-1:1;
}

struct point
{
double x,y;
} p[_max],res[_max];

bool mult(point sp,point ep,point op)
{
return (sp.x - op.x) * (ep.y - op.y)
>= (ep.x - op.x) * (sp.y - op.y);
}

bool operator < (const point &l, const point &r)
{
return l.y < r.y ||(l.y == r.y && l.x < r.x);
}

int graham(point pnt[],int n, point res[]) //构造凸包
{
int i,len ,k = 0,top = 1;
sort(pnt,pnt+n);
if(n == 0)return 0;
res[0] = pnt[0];
if(n == 1)return 1;
res[1] = pnt[1];
if(n == 2)return 2;
res[2] = pnt[2];
for(int i =2; i < n; ++ i)
{
while(top && mult(pnt[i],res[top],res[top-1]))
top--;
res[++top] = pnt[i];
}
len = top;
res[++top] = pnt[n - 2];
for(i = n - 3; i >= 0; -- i)
{
while(top!=len && mult(pnt[i],res[top],res[top-1]))
top--;
res[++top] = pnt[i];
}
return top;//返回凸包中点的个数
}

double len(point A,point B) //返回向量AB的模平方
{
return (A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y);
}

double dot(point A,point B,point C) //点乘
{
return (C.x-A.x)*(B.x-A.x)+(C.y-A.y)*(B.y-A.y);
}

double cross(point A,point B,point C) //叉乘
{
return (B.x-A.x)*(C.y-A.y)-(B.y-A.y)*(C.x-A.x);
}

double minRetangleCover() //最小矩形面积覆盖(旋转卡壳)
{
if(n < 3) return 0.0;
res
= res [0];
double ans = -1;
int r = 1, p = 1,q;
for(int i = 0; i < n; ++ i)
{
//卡出离边 res[i]-res[i+1]最远的点
while(sgn(cross(res[i],res[i+1],res[r+1])-cross(res[i],res[i+1],res[r]))>=0)
r = (r+1)% n;
//卡出res[i]-res[i+1]方向上正向n最远的点
while(sgn(dot(res[i],res[i+1],res[p+1])-dot(res[i],res[i+1],res[p]))>=0)
p = (p+1)% n;
if(i == 0) q = p;
//卡出res[i]-res[i+1]方向上负向最远的点
while(sgn(dot(res[i],res[i+1],res[q+1])-dot(res[i],res[i+1],res[q]))<=0)
q = (q+1)% n;
double d = len(res[i],res[i+1]);
double temp = cross(res[i],res[i+1],res[r])*(dot(res[i],res[i+1],res[p])-dot(res[i],res[i+1],res[q]))/d;
if(ans < 0 || ans > temp) ans = temp;
}
return ans;
}

int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
while(scanf("%d",&n)==1 && n)
{
for(int i = 0; i < n; ++ i)
{
scanf("%lf%lf",&p[i].x,&p[i].y);
}
n = graham(p,n,res);//构造凸包
printf("%.4f\n",minRetangleCover());
}
return 0;
}