BZOJ3451: Tyvj1953 Normal

  1 #include<cstdio>
  2 
  3 #include<cstdlib>
  4 
  5 #include<cmath>
  6 
  7 #include<cstring>
  8 
  9 #include<algorithm>
 10 
 11 #include<iostream>
 12 
 13 #include<vector>
 14 
 15 #include<map>
 16 
 17 #include<set>
 18 
 19 #include<queue>
 20 
 21 #include<string>
 22 
 23 #define inf 1000000000
 24 
 25 #define maxn 50000+5
 26 
 27 #define maxm 20000000+5
 28 
 29 #define eps 1e-10
 30 
 31 #define ll long long
 32 
 33 #define pa pair<int,int>
 34 
 35 #define for0(i,n) for(int i=0;i<=(n);i++)
 36 
 37 #define for1(i,n) for(int i=1;i<=(n);i++)
 38 
 39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 40 
 41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 42 #define for4(i,x) for(int i=head[x],y;i;i=e[i].next)
 43 
 44 #define mod 1000000007
 45 
 46 using namespace std;
 47 
 48 inline int read()
 49 
 50 {
 51 
 52     int x=0,f=1;char ch=getchar();
 53 
 54     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
 55 
 56     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
 57 
 58     return x*f;
 59 
 60 }
 61 int n,mx,m,len,cnt,sum,tot,rt,rev[maxn],head[maxn],d[maxn],f[maxn],s[maxn],g[maxn];
 62 bool del[maxn];
 63 ll ret[maxn];
 64 const double PI=acos(-1.0);
 65 struct cp
 66 {
 67     double x,y;
 68     cp operator +(cp b){return (cp){x+b.x,y+b.y};}
 69     cp operator -(cp b){return (cp){x-b.x,y-b.y};}
 70     cp operator *(cp b){return (cp){x*b.x-y*b.y,x*b.y+y*b.x};}
 71 };
 72 cp a[2*maxn],b[2*maxn],c[2*maxn],y[2*maxn];
 73 struct edge{int go,next;}e[2*maxn];
 74 inline void insert(int x,int y)
 75 {
 76     e[++tot]=(edge){y,head[x]};head[x]=tot;
 77     e[++tot]=(edge){x,head[y]};head[y]=tot;
 78 }
 79 inline void getrt(int x,int fa)
 80 {
 81     s[x]=1;f[x]=0;
 82     for4(i,x)if(!del[y=e[i].go]&&y!=fa)
 83     {
 84         getrt(y,x);
 85         s[x]+=s[y];
 86         f[x]=max(f[x],s[y]);
 87     }
 88     f[x]=max(f[x],sum-f[x]);
 89     if(f[x]<f[rt])rt=x;
 90 }
 91 inline void getdep(int x,int fa,int w)
 92 {
 93     if(w>mx)mx=w;
 94     d[++cnt]=w;
 95     for4(i,x)if(!del[y=e[i].go]&&y!=fa)getdep(y,x,w+1);
 96 }
 97 inline void get(int n)
 98 {
 99     n++;n=2*n-1;
100     m=1,len=0;
101     while(m<=n)m<<=1,len++;
102     for0(i,m-1)
103     {
104         int x=i,y=0;
105         for1(j,len)y<<=1,y|=x&1,x>>=1;
106         rev[i]=y;
107     }
108 }
109 inline void fft(cp *x,int n,int flag)
110 {
111     for0(i,n-1)y[rev[i]]=x[i];
112     for0(i,n-1)x[i]=y[i];
113     for(int m=2;m<=n;m<<=1)
114     {
115         cp wn=(cp){cos(2.0*PI*flag/m),sin(2.0*PI*flag/m)};
116         for(int i=0;i<n;i+=m)
117         {
118             cp w=(cp){1,0};int mid=m>>1;
119             for0(j,mid-1)
120             {
121                 cp u=x[i+j],v=x[i+j+mid]*w;
122                 x[i+j]=u+v;x[i+j+mid]=u-v;
123                 w=w*wn;
124             }
125         }
126     }
127     if(flag==-1)for0(i,n-1)x[i].x/=n;
128 }
129 inline void work(int x)
130 {
131     //cout<<"XXXXX"<<‘ ‘<<x<<‘ ‘<<"XXXX"<<endl;
132     del[x]=1;mx=0;
133     for4(i,x)if(!del[y=e[i].go])
134     {
135         cnt=0;
136         getdep(y,x,1);get(mx);
137         for0(j,m-1)a[j]=(cp){g[j],0},b[j]=(cp){0,0};
138         for1(j,cnt)b[d[j]].x+=1,g[d[j]]++;
139         //for0(j,m-1)cout<<j<<‘ ‘<<a[j].x<<‘ ‘<<b[j].x<<endl;
140         fft(a,m,1);fft(b,m,1);
141         for0(j,m-1)c[j]=a[j]*b[j];
142         fft(c,m,-1);
143         for0(j,m-1)ret[j]+=c[j].x+0.5;
144         //for0(j,m-1)cout<<j<<‘ ‘<<c[j].x<<‘ ‘<<c[j].y<<endl;
145     }
146     for1(i,mx)ret[i]+=g[i],g[i]=0;
147     for4(i,x)if(!del[y=e[i].go])
148     {
149         sum=s[y];rt=0;
150         getrt(y,0);
151         work(rt);
152     }
153 }    
154 
155 int main()
156 
157 {
158 
159     freopen("input.txt","r",stdin);
160 
161     freopen("output.txt","w",stdout);
162 
163     n=read();
164     for1(i,n-1)insert(read()+1,read()+1);
165     sum=n;f[rt=0]=inf;
166     getrt(1,0);
167     work(rt);
168     double ans=0.0;
169     for1(i,n)
170     ans+=(double)ret[i]/(double)(i+1);//cout<<i<<‘ ‘<<ret[i]<<endl;
171     printf("%.4f\n",n+2*ans);
172 
173     return 0;
174 
175 }