bzoj 1185: [HNOI2007]最小矩形覆盖
%%hzwer，，可以通过蹩脚的画图法发现，矩形的一边会和凸包一边重合。所以就可以枚举边，然后在用旋转卡壳的方法求出左边界，上边界，右边界就是当前的最小面积。
这题细节太神了、、、（应该先做下一题，，，）
1 #include<cstdio> 2 #include<algorithm> 3 #include<cmath> 4 #define eps 1e-8 5 using namespace std; 6 double ans=1e60; 7 int n,top; 8 struct point{ 9 double x,y; 10 point(){} 11 point(double _x, double _y):x(_x),y(_y){} 12 friend bool operator < (point a, point b){ 13 return fabs(a.y-b.y)<eps?a.x<b.x:a.y<b.y; 14 } 15 friend bool operator == (point a, point b){ 16 return fabs(a.x-b.x)<eps && fabs(a.y-b.y)<eps; 17 } 18 friend bool operator != (point a, point b){ 19 return !(a==b); 20 } 21 friend point operator + (point a, point b){ 22 return point(a.x+b.x,a.y+b.y); 23 } 24 friend point operator - (point a, point b){ 25 return point (a.x-b.x,a.y-b.y); 26 } 27 friend point operator * (point a, double b){ 28 return point (a.x*b,a.y*b); 29 } 30 friend double operator * (point a, point b){ 31 return a.x*b.y-a.y*b.x; 32 } 33 friend double operator / (point a, point b){ 34 return a.x*b.x+a.y*b.y; 35 } 36 friend double dis(point a){ 37 return sqrt(a.x*a.x+a.y*a.y); 38 } 39 }p[50005],q[50005],t[5]; 40 bool cmp(point a, point b) 41 { 42 double t=(a-p[1])*(b-p[1]); 43 if (fabs(t)<eps) return dis(p[1]-a)-dis(p[1]-b)<0; 44 return t>0; 45 } 46 void Graham() 47 { 48 for (int i=2; i<=n; i++) 49 if (p[i]<p[1]) swap(p[1],p[i]); 50 sort(p+2,p+n+1,cmp); 51 q[++top]=p[1]; 52 for (int i=2; i<=n; i++) 53 { 54 while (top>1 && (q[top]-q[top-1])*(p[i]-q[top])<eps) top--; 55 q[++top]=p[i]; 56 } 57 q[0]=q[top]; 58 } 59 void RC() 60 { 61 int l=1,r=1,p=1; 62 double L,R,D,H; 63 for (int i=0; i<top; i++) 64 { 65 D=dis(q[i]-q[i+1]); 66 while ((q[i+1]-q[i])*(q[p+1]-q[i])-(q[i+1]-q[i])*(q[p]-q[i])>-eps) p=(p+1)%top; 67 while ((q[i+1]-q[i])/(q[r+1]-q[i])-(q[i+1]-q[i])/(q[r]-q[i])>-eps) r=(r+1)%top; 68 if (i==0) l=r; 69 while ((q[i+1]-q[i])/(q[l+1]-q[i])-(q[i+1]-q[i])/(q[l]-q[i])<eps) l=(l+1)%top; 70 L=(q[i+1]-q[i])/(q[l]-q[i])/D; 71 R=(q[i+1]-q[i])/(q[r]-q[i])/D; 72 H=(q[i+1]-q[i])*(q[p]-q[i])/D; 73 double tmp=fabs(H)*(R-L); 74 if (tmp<ans) 75 { 76 ans=tmp; 77 t[0]=q[i]+(q[i+1]-q[i])*(R/D); 78 t[1]=t[0]+(q[r]-t[0])*(H/dis(t[0]-q[r])); 79 t[2]=t[1]-(t[0]-q[i])*((R-L)/dis(q[i]-t[0])); 80 t[3]=t[2]-(t[1]-t[0]); 81 } 82 } 83 } 84 int main() 85 { 86 scanf("%d",&n); 87 for (int i=1; i<=n; i++) scanf("%lf%lf",&p[i].x,&p[i].y); 88 Graham(); RC(); 89 printf("%.5lf\n",ans); 90 int fir=0; 91 for (int i=1; i<=3; i++) 92 if (t[i]<t[fir]) fir=i; 93 for (int i=0; i<=3; i++) 94 printf("%.5lf %.5lf\n",t[(i+fir)%4].x,t[(i+fir)%4].y); 95 return 0; 96 }
