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();
}