BZOJ1337: 最小圆覆盖
题目：求n个点的最小圆覆盖。
题解：最小圆覆盖，上模板。复杂度证明可以戳：这里
代码：
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set> 10 #include<queue> 11 #include<string> 12 #define inf 1000000000 13 #define maxn 1000000+5 14 #define maxm 200000+5 15 #define eps 1e-6 16 #define ll long long 17 #define ull unsigned long long 18 #define pa pair<int,int> 19 #define for0(i,n) for(int i=0;i<=(n);i++) 20 #define for1(i,n) for(int i=1;i<=(n);i++) 21 #define for2(i,x,y) for(int i=(x);i<=(y);i++) 22 #define for3(i,x,y) for(int i=(x);i>=(y);i--) 23 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go) 24 #define for5(n,m) for(int i=1;i<=n;i++)for(int j=1;j<=m;j++) 25 #define mod 1000000007 26 #define lch k<<1,l,mid 27 #define rch k<<1|1,mid+1,r 28 #define sqr(x) (x)*(x) 29 #define db double 30 using namespace std; 31 inline int read() 32 { 33 int x=0,f=1;char ch=getchar(); 34 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 35 while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();} 36 return x*f; 37 } 38 struct point 39 { 40 db x,y; 41 point operator +(point b){return (point){x+b.x,y+b.y};} 42 point operator /(db b){return (point){x/b,y/b};} 43 }a[maxn]; 44 int n; 45 db ans; 46 inline db dist(point a,point b){return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));} 47 inline point calc(db a,db b,db c,db d,db e,db f) 48 { 49 return (point){(c*e-f*b)/(a*e-d*b),(a*f-c*d)/(a*e-d*b)}; 50 } 51 inline point get(point a,point b,point c) 52 { 53 return calc(b.x-a.x,b.y-a.y,(sqr(b.x)+sqr(b.y)-sqr(a.x)-sqr(a.y))/2.0, 54 c.x-b.x,c.y-b.y,(sqr(c.x)+sqr(c.y)-sqr(b.x)-sqr(b.y))/2.0); 55 } 56 int main() 57 { 58 freopen("input.txt","r",stdin); 59 freopen("output.txt","w",stdout); 60 n=read(); 61 for1(i,n)scanf("%lf%lf",&a[i].x,&a[i].y); 62 for1(i,n)swap(a[rand()%n+1],a[rand()%n+1]); 63 ans=0; 64 for1(i,n)if(dist(a[i],a[0])>ans+eps) 65 { 66 a[0]=a[i];ans=0; 67 for1(j,i-1)if(dist(a[j],a[0])>ans+eps) 68 { 69 a[0]=(a[i]+a[j])/2;ans=dist(a[0],a[i]); 70 for1(k,j-1)if(dist(a[k],a[0])>ans+eps) 71 { 72 a[0]=get(a[i],a[j],a[k]);ans=dist(a[0],a[i]); 73 } 74 } 75 } 76 printf("%.2f %.2f %.2f\n",a[0].x,a[0].y,ans); 77 return 0; 78 }
我不会说告诉你这题是三倍经验的
