BZOJ1185 HNOI2007 最小矩阵覆盖

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define eps 1e-8
 4 using namespace std;
 5 inline int read(){
 6     int x=0;int f=1;char ch=getchar();
 7     while(!isdigit(ch)) {if(ch==‘-‘) f-=-1;ch=getchar();}
 8     while(isdigit(ch)) {x=x*10+ch-‘0‘;ch=getchar();}
 9     return x*f;
10 }
11 const int MAXN=1e6+10;
12 struct node{
13     double x,y;
14 }s[MAXN],p[MAXN],t[10];
15 int top,n;
16 double ans=1e60;
17 inline  bool operator < (node n,node m){
18     return abs(n.y-m.y)<eps?n.x<m.x:n.y<m.y;
19 }
20 inline node operator + (node n,node m){
21     node t;t.x=n.x+m.x;t.y=n.y+m.y;return t;
22 }
23 inline node operator - (node n,node m){
24     node t;t.x=n.x-m.x;t.y=n.y-m.y;return t;
25 }
26 inline double operator * (node n,node m){
27     return n.x*m.y-n.y*m.x;
28 }
29 inline node operator * (node n,double m){
30     node t;t.x=n.x*m;t.y=n.y*m;return t;
31 }
32 inline int dcmp(node n,node m){
33     return n.x-m.x<eps&&n.y-m.y<eps;
34 }
35 inline double dis(node n){
36     return sqrt(n.x*n.x+n.y*n.y);
37 }
38 inline bool mycmp(node n,node m){
39     double t=(n-p[1])*(m-p[1]);
40     if(abs(t)<eps) return dis(n-p[1])-dis(m-p[1])<0;
41     else return t>0;
42 }
43 inline double operator / (node n,node m){//dianji
44     return n.x*m.x+n.y*m.y;
45 }
46 void graham(){
47     for(int i=2;i<=n;i++){
48         if(p[i]<p[1]) swap(p[i],p[1]);
49     }    
50     sort(p+2,p+n+1,mycmp);
51     s[++top]=p[1];
52     for(int i=2;i<=n;i++){
53         while(top>1&&(s[top]-s[top-1])*(p[i]-s[top])<eps) top--;
54         s[++top]=p[i];
55     }
56     s[0]=s[top];
57 }
58 void RC(){
59     int l=1,r=1,p=1;
60     double R,L,D,H;
61     for(int i=0;i<top;i++){
62         D=dis(s[i]-s[i+1]);
63         while((s[i+1]-s[i])*(s[p+1]-s[i])-(s[i+1]-s[i])*(s[p]-s[i])>-eps) p=(p+1)%top;
64         while((s[i+1]-s[i])/(s[r+1]-s[i])-(s[i+1]-s[i])/(s[r]-s[i])>-eps) r=(r+1)%top;
65         if(i==0) l=r;
66         while((s[i+1]-s[i])/(s[l+1]-s[i])-(s[i+1]-s[i])/(s[l]-s[i])<eps) l=(l+1)%top;
67         L=(s[i+1]-s[i])/(s[l]-s[i])/D;R=(s[i+1]-s[i])/(s[r]-s[i])/D;
68         H=(s[i+1]-s[i])*(s[p]-s[i])/D;
69         if(H<0) H=-H;
70         //cout<<l<<‘ ‘<<r<<endl;
71         double tmp=(R-L)*H;
72         if(tmp<ans){
73             ans=tmp;
74             t[0]=s[i]+(s[i+1]-s[i])*(R/D);
75             t[1]=t[0]+(s[r]-t[0])*(H/dis(t[0]-s[r]));
76             t[2]=t[1]-(t[0]-s[i])*((R-L)/dis(s[i]-t[0]));
77             t[3]=t[2]-(t[1]-t[0]);
78         }
79     }
80 }
81 int main(){
82     //freopen("All.in","r",stdin);
83     //freopen("a.out","w",stdout);
84     n=read();
85     for(int i=1;i<=n;i++){
86         scanf("%lf%lf",&p[i].x,&p[i].y);
87     }
88     graham();
89     RC();
90     printf("%.5lf\n",ans);
91     int fir=0;
92     for(int i=1;i<=3;i++){
93         if(t[i]<t[fir]) fir=i;
94     }
95     for(int i=0;i<=3;i++){
96         printf("%.5lf %.5lf\n",t[(i+fir)%4].x,t[(i+fir)%4].y);
97     }
98     return 0;
99 }