判断矩形能包围点集的最小面积：凸包
1 #include <iostream> 2 #include <cmath> 3 #include <cstdio> 4 #include <cstring> 5 #include <cstdlib> 6 #include <string> 7 #include <sstream> 8 #include <algorithm> 9 #define Max 2147483647 10 #define INF 0x7fffffff 11 #define ll long long 12 #define mem(a,b) memset(a,b,sizeof(a)) 13 #define repu(i, a, b) for(int i = (a); i < (b); i++) 14 const double PI=-acos(-1.0); 15 #define eps 1e-8 16 #define N 50010 17 using namespace std; 18 struct Point 19 { 20 double x,y; 21 Point() {} 22 Point(double x0,double y0):x(x0),y(y0) {} 23 }; 24 Point p[N]; 25 int con[N]; 26 int cn; 27 int n; 28 struct Line 29 { 30 Point a,b; 31 Line() {} 32 Line(Point a0,Point b0):a(a0),b(b0) {} 33 }; 34 double Xmult(Point o,Point a,Point b) 35 { 36 return (a.x-o.x)*(b.y-o.y)-(b.x-o.x)*(a.y-o.y); 37 } 38 double Dmult(Point o,Point a,Point b) 39 { 40 return (a.x-o.x)*(b.x-o.x)+(a.y-o.y)*(b.y-o.y); 41 } 42 int Sig(double a) 43 { 44 return a<-eps?-1:a>eps; 45 } 46 double Dis(Point a,Point b) 47 { 48 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 49 } 50 int cmp(Point a,Point b) 51 { 52 double d=Xmult(p[0],a,b); 53 if(d>0) 54 return 1; 55 if(d==0 && Dis(p[0],a)<Dis(p[0],b)) 56 return 1; 57 return 0; 58 } 59 double min(double a,double b) 60 { 61 return a<b?a:b; 62 } 63 void Graham() 64 { 65 int i,ind=0; 66 for(i=1; i<n; i++) 67 if(p[ind].y>p[i].y || (p[ind].y==p[i].y) && p[ind].x>p[i].x) 68 ind=i; 69 swap(p[ind],p[0]); 70 sort(p+1,p+n,cmp); 71 con[0]=0; 72 con[1]=1; 73 cn=1; 74 for(i=2; i<n; i++) 75 { 76 while(cn>0 && Sig(Xmult(p[con[cn-1]],p[con[cn]],p[i]))<=0) 77 cn--; 78 con[++cn]=i; 79 } 80 int tmp=cn; 81 for(i=n-2; i>=0; i--) 82 { 83 while(cn>tmp && Sig(Xmult(p[con[cn-1]],p[con[cn]],p[i]))<=0) 84 cn--; 85 con[++cn]=i; 86 } 87 } 88 double Solve() 89 { 90 int t,r,l; 91 double ans=999999999; 92 t=r=1; 93 if(cn<3) 94 return 0; 95 for(int i=0; i<cn; i++) 96 { 97 while(Sig( Xmult(p[con[i]],p[con[i+1]],p[con[t+1]])- 98 Xmult(p[con[i]],p[con[i+1]],p[con[t]]) )>0) 99 t=(t+1)%cn; 100 while(Sig( Dmult(p[con[i]],p[con[i+1]],p[con[r+1]])- 101 Dmult(p[con[i]],p[con[i+1]],p[con[r]]) )>0) 102 r=(r+1)%cn; 103 if(!i) l=r; 104 while(Sig( Dmult(p[con[i]],p[con[i+1]],p[con[l+1]])- 105 Dmult(p[con[i]],p[con[i+1]],p[con[l]]) )<=0) 106 l=(l+1)%cn; 107 double d=Dis(p[con[i]],p[con[i+1]]); 108 double tmp=Xmult(p[con[i]],p[con[i+1]],p[con[t]])* 109 ( Dmult(p[con[i]],p[con[i+1]],p[con[r]])- 110 Dmult(p[con[i]],p[con[i+1]],p[con[l]]) )/d/d; 111 ans=min(ans,tmp); 112 } 113 return ans; 114 } 115 int main() 116 { 117 int i,T; 118 scanf("%d",&T); 119 repu(kase,1,T+1) 120 { 121 scanf("%d",&n); 122 n *= 4; 123 for(i=0; i<n; i++) 124 scanf("%lf%lf",&p[i].x,&p[i].y); 125 Graham(); 126 printf("Case #%d: %.0lf ",kase,Solve()); 127 } 128 return 0; 129 }