BZOJ 2878 迷失游乐园

http://www.lydsy.com/JudgeOnline/problem.php?id=2878

题意:n个点的图，保证图联通，有n-1或者n条边，求从任意一个点出发，不经过相同点，最终无路可走的路径长度期望。

思路:m=n-1时，可以用dp，处理出i点往下，i点往上走的路径长度期望，然后O(n)统计就可以了。

m=n时，由于环上点点数不超过20，那这就是基环树，我们考虑枚举基环树上面的点i，假设我们到了i，并走向了i的子树。

这样我们就把环断开了，然后我们再两遍顺逆走这个环，就可以处理出up数组了，然后再分别下传到每个子树里。

 #include<algorithm>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<iostream>
 int flag,n,m,f[],len[],vis[],tot,go[],next[],cnt,cir[],huan[],first[],val[],du[];
 double down[],up[];
 int read(){
     char ch=getchar();int t=,f=;
     while (ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}
     while (''<=ch&&ch<=''){t=t*+ch-'';ch=getchar();}
     return t*f;
 }
 void insert(int x,int y,int z){
     tot++;
     go[tot]=y;
     next[tot]=first[x];
     first[x]=tot;
     val[tot]=z;
 }
 void add(int x,int y,int z){
     insert(x,y,z);insert(y,x,z);
 }
 void dfs1(int x,int fa){
     vis[x]=;
     for (int i=first[x];i;i=next[i]){
         int pur=go[i];
         if (vis[pur]) continue;
         dfs1(pur,x);
         down[x]+=(double)val[i]+(((double)down[pur])/((double)(du[pur]-+(du[pur]==))));
     }
 }
 void dfs2(int x,int fa){
     if (!cir[x]&&fa!=) up[x]=len[x]+((double)((double)up[fa]+down[fa]-len[x]-((double)down[x])/((double)du[x]-+(du[x]==))))/((double)(du[fa]-+(du[fa]==)));
     vis[x]=;
     for (int i=first[x];i;i=next[i]){
         int pur=go[i];
         if (vis[pur]) continue;
         f[pur]=x;
         len[pur]=val[i];
         dfs2(pur,x);
     }
 }
 void work1(){
     dfs1(,);
     for (int i=;i<=n;i++) vis[i]=;
     dfs2(,);
     double ans=;
     for (int i=;i<=n;i++) ans+=((double)down[i]+(double)up[i])/((double)du[i]);
     ans/=((double)(n));
     printf("%.5f\n",ans);
 }
 void dfs(int x,int fa){
     vis[x]=;
     for (int i=first[x];i;i=next[i]){
         int pur=go[i];
         if (flag) return;
         if (f[x]!=pur&&vis[pur]){
             len[pur]=val[i];
             int j=x;
             while (j!=f[pur]){
                 cnt++;
                 cir[j]=;
                 huan[cnt]=j;
                 j=f[j];
             }
             flag=;
             return;
         }else if (!vis[pur]){
             len[pur]=val[i];
             f[pur]=x;
             dfs(pur,x);
         }
     }
 }
 void work2(){
      flag=;
      dfs(,);cnt++;
      for (int i=;i<=n;i++) vis[i]=;
      for (int i=;i<cnt;i++) vis[huan[i]]=;
      for (int i=;i<cnt;i++) dfs1(huan[i],);
      for (int i=;i<cnt;i++){
             int x=(i+)%cnt;double s=;
             while (x!=i){
                 if ((x+)%cnt==i) up[huan[i]]+=(double)s*len[huan[(x-+cnt)%cnt]]+((double)(s*down[huan[x]]))/((double)(du[huan[x]]-+(du[huan[x]]==)));
                 else up[huan[i]]+=(double)s*len[huan[(x-+cnt)%cnt]]+((double)(s*down[huan[x]]))/((double)(du[huan[x]]-));
                 s=s/((double)(du[huan[x]]-));
                 x=(x+)%cnt;
             }
             x=(i-+cnt)%cnt;s=;
             while (x!=i){
                 if ((x-+cnt)%cnt==i) up[huan[i]]+=(double)s*len[huan[x]]+((double)(s*down[huan[x]]))/((double)(du[huan[x]]-+(du[huan[x]]==)));
                 else up[huan[i]]+=(double)s*len[huan[(x)%cnt]]+((double)(s*down[huan[x]]))/((double)(du[huan[x]]-));
                 s=s/((double)(du[huan[x]]-));
                 x=(x-+cnt)%cnt;
             }
     }
     for (int i=;i<=n;i++) vis[i]=;
     for (int i=;i<cnt;i++) vis[huan[i]]=;
     for (int i=;i<cnt;i++) dfs2(huan[i],);
     double ans=;
     for (int i=;i<=n;i++) ans+=(down[i]+up[i])/(du[i]);
     printf("%.5f\n",ans/n);
 }
 int main(){
     cnt=-;
     n=read();m=read();flag=;
     for (int i=;i<=m;i++){
         int x=read(),y=read(),z=read();
         add(x,y,z);
         du[x]++;du[y]++;
     }
     if (m==n-) work1();
     else work2();
 }