UVA 10173 旋转卡壳

题意：
给出一些点，求最小的覆盖这些点的矩形的面积。

题解：

枚举下边界（是一条边），然后暴力卡壳左右边界（点），再暴力上边界（点），更新答案。

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

#define N 2222
#define EPS 1e-7
#define INF 1e20

using namespace std;

struct PO
{
    double x,y;
}p[N],stk[N],o;

int n,top;
int s[10];
double ans;

inline int dc(double x)
{
    if(x>EPS) return 1;
    else if(x<-EPS) return -1;
    return 0;
}

inline bool cmp(const PO &a,const PO &b)
{
    if(dc(a.x-b.x)==0) return a.y<b.y;
    return a.x<b.x;
}

inline PO operator +(PO a,PO b)
{
    PO c;
    c.x=a.x+b.x;
    c.y=a.y+b.y;
    return c;
}

inline PO operator -(PO a,PO b)
{
    PO c;
    c.x=a.x-b.x;
    c.y=a.y-b.y;
    return c;
}

inline double cross(PO &a,PO &b,PO &c)
{
    return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
}

inline double getangle(PO &a,PO &b,PO&c,PO &d)
{
    PO t=c+(a-d);
    return cross(b,a,t);
}

inline PO getfline(PO &a,PO &b,PO &c)//得到垂线 
{
    PO d=c-b,e;
    e.x=a.x-d.y;
    e.y=a.y+d.x;
    return e;
}

inline double getdis(PO &a,PO &b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

inline double getdis_ps(PO &a,PO &b,PO &c)
{
    return fabs(cross(a,b,c))/getdis(b,c);
}

inline void getside()
{
    for(int i=0;i<4;i++) s[i]=0;
    for(int i=1;i<top;i++)
    {
        if(dc(stk[i].y-stk[s[0]].y)<0) s[0]=i;
        if(dc(stk[i].x-stk[s[1]].x)<0) s[1]=i;
        if(dc(stk[i].y-stk[s[2]].y)>0) s[2]=i;
        if(dc(stk[i].x-stk[s[3]].x)>0) s[3]=i;
    }
    // 0 == ymin, 1 == xmin, 2 == ymax ,3 == xmax;
}

inline void rotating_calipers()
{
    getside();
    int tmp=s[0];
    ans=INF;
    do
    {//枚举下边界（直线） 
        PO t=getfline(stk[s[0]],stk[s[0]],stk[(s[0]+1)%top]);
        while(dc(getangle(t,stk[s[0]],stk[s[1]],stk[(s[1]+1)%top]))<0) s[1]=(s[1]+1)%top;//卡右边界 
        while(dc(getangle(stk[s[0]],t,stk[s[3]],stk[(s[3]+1)%top]))<0) s[3]=(s[3]+1)%top;//卡做边界 
        while(dc(getdis_ps(stk[(s[2]+1)%top],stk[s[0]],stk[(s[0]+1)%top])-
                getdis_ps(stk[s[2]],stk[s[0]],stk[(s[0]+1)%top]))>0) s[2]=(s[2]+1)%top;//卡上边界 
        double a=getdis_ps(stk[s[2]],stk[s[0]],stk[(s[0]+1)%top]);
        t=getfline(stk[s[3]],stk[s[0]],stk[(s[0]+1)%top]);
        double b=getdis_ps(stk[s[1]],stk[s[3]],t);
        ans=min(ans,a*b);
        s[0]=(s[0]+1)%top;
    }while(s[0]!=tmp);
}

inline void graham()
{
    sort(p+1,p+1+n,cmp);
    top=-1;
    stk[++top]=p[1]; stk[++top]=p[2];
    for(int i=3;i<=n;i++)
    {
        while(top>=1&&dc(cross(stk[top-1],stk[top],p[i]))<=0) top--;
        stk[++top]=p[i];
    }
    int tmp=top;
    for(int i=n-1;i>=1;i--)
    {
        while(top>=tmp+1&&dc(cross(stk[top-1],stk[top],p[i]))<=0) top--;
        stk[++top]=p[i];
    }
}

inline void read()
{
    for(int i=1;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);
}

inline void go()
{
    if(n<=2) ans=0.0;
    else graham(),rotating_calipers();
    printf("%.4lf\n",ans);
}

int main()
{
    while(~scanf("%d",&n)&&n) read(),go();
    return 0;
}

这题傻X了，凸包写错了。。查了好久，都改得和题解一样了。。。~~~~(>_<)~~~~ 

一下自己yy的。

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

#define N 2020
#define EPS 1e-7

using namespace std;

struct PO
{
    double x,y;
    inline void prt() {printf("%lf     %lf\n",x,y);}
}p[N],stk[N],res[N],o;

int n,tot;

inline void read()
{
    for(int i=1;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);
}

inline int dc(double x)
{
    if(x>EPS) return 1;
    else if(x<-EPS) return -1;
    return 0;
}

inline bool cmp(const PO &a,const PO &b)
{
    if(dc(a.x-b.x)==0) return a.y<b.y;
    return a.x<b.x;
}

inline PO operator +(PO a,PO b)
{
    a.x+=b.x; a.y+=b.y;
    return a;
}

inline PO operator -(PO a,PO b)
{
    a.x-=b.x; a.y-=b.y;
    return a;
}

inline double cross(PO &a,PO &b,PO &c)
{
    return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
}

inline double dot(PO &a,PO &b,PO &c)
{
    return (b.x-a.x)*(c.x-a.x)+(b.y-a.y)*(c.y-a.y);
}

inline double getdis(PO &a,PO &b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

inline void graham()
{
    sort(p+1,p+1+n,cmp);
    tot=-1; int top=0;
    for(int i=1;i<=n;i++)
    {
        while(top>=2&&dc(cross(stk[top-1],stk[top],p[i]))<=0) top--;
        stk[++top]=p[i];
    }
    for(int i=1;i<=top;i++) res[++tot]=stk[i];
    top=0;
    for(int i=n;i>=1;i--)
    {
        while(top>=2&&dc(cross(stk[top-1],stk[top],p[i]))<=0) top--;
        stk[++top]=p[i];
    }
    for(int i=2;i<=top;i++) res[++tot]=stk[i];
}

inline int getangle_dot(PO &a,PO &b,PO &c,PO &d)
{
    PO e=d-(c-a);
    return dc(dot(a,b,e));
}

inline double getangle_cross(PO &a,PO &b,PO &c,PO &d)
{
    PO e=d-(c-a);
    return dc(cross(a,b,e));
}

inline double getlen(PO &a)
{
    return sqrt(a.x*a.x+a.y*a.y);
}

inline double getty(PO &a,PO &b)
{
    return dot(o,a,b)/getlen(b);
}

inline void rotating_calipers()
{
    double ans=1e20;
    PO sa,sb;
    for(int i=0;i<tot;i++)
    {
        int rt=(i+1)%tot;
        while(getangle_dot(res[i],res[(i+1)%tot],res[rt],res[(rt+1)%tot])>0) rt=(rt+1)%tot;
        int lt=i;
        while(getangle_dot(res[i],res[(i+1)%tot],res[(lt-1+tot)%tot],res[lt])>0) lt=(lt-1+tot)%tot;
        int usp=(i+1)%tot;
        while(getangle_cross(res[i],res[i+1],res[usp],res[(usp+1)%tot])>0) usp=(usp+1)%tot;
        double h=fabs(cross(res[i],res[i+1],res[usp]))/getdis(res[i],res[i+1]);
        sa=res[rt]-res[lt]; sb=res[i+1]-res[i];
        ans=min(ans,fabs(h*getty(sa,sb)));
    }
    printf("%.4lf\n",ans);
}

inline void go()
{
    graham();
    if(tot<=2) printf("0.0000\n");
    else if(tot==3)  printf("%.4lf\n",fabs(cross(res[0],res[1],res[2])));
    else rotating_calipers();
}

int main()
{
    while(scanf("%d",&n),n)read(),go();
    return 0;
}