laekov

水水的dp。拿pascal玩玩边表。然后莫名其妙地发现在main上先是ce然后全wa。下到数据一测就过了，交bzoj还是过了。不开心。

必定是一个中间点然后连三条出去。所以直接枚举中间点然后bfs就好了。相当于每条边被转移了两次所以还是O(n^2)的。最初想错了想成三方了把自己吓了一跳。

pascal啊。

type

edge = ^edger;

edger = record

t: longint;

next: edge;

end;

const

maxn = 5009;

//ONLINE_JUDGE = false;

ONLINE_JUDGE = true;

var

n, i, j, u, v: longint;

ans: int64;

d, cd, sd, td: array [0 .. maxn] of longint;

head: array [0 .. maxn] of edge;

ei: edge;

procedure addEdge(u, v: longint);

var

ep: edge;

begin

new(ep);

ep^. t := v;

ep^. next := head[u];

head[u] := ep;

end;

procedure bfs(p0: longint);

var

hd, tl, p: longint;

q: array [0 .. maxn] of longint;

e: edge;

begin

fillchar(cd, sizeof(cd), 0);

hd := 0;

tl := 1;

q[0] := p0;

d[p0] := 2;

while hd < tl do begin

p := q[hd];

inc(hd);

inc(cd[d[p]]);

e := head[p];

while e <> NIL do begin

if d[e^. t] = 0 then begin

d[e^. t] := d[p] + 1;

q[tl] := e^. t;

inc(tl);

end;

e := e^. next;

end;

end;

end;

begin

if not ONLINE_JUDGE then begin

assign(input, 'in.txt');

reset(input);

end;

readln(n);

for i := 2 to n do begin

readln(u, v);

addEdge(u, v);

addEdge(v, u);

end;

ans := 0;

for i := 1 to n do begin

fillchar(d, sizeof(d), 0);

fillchar(sd, sizeof(sd), 0);

fillchar(td, sizeof(td), 0);

d[i] := 1;

ei := head[i];

while ei <> nil do begin

bfs(ei^. t);

for j := 2 to n do begin

inc(ans, td[j] * cd[j]);

inc(td[j], sd[j] * cd[j]);

inc(sd[j], cd[j]);

end;

ei := ei^. next;

end;

end;

writeln(ans);

close(input);

end.