找出一条直径，以直径上的每个点为开头或结尾，分别求它的ECC即可
可以证明的是，如果图中存在多条路径，则在任何一条直径上都存在一条core（反证法，用到直径的距离最大性）。因此只需讨论一条直径上的core的情况即可。
program core;
type
node=record
x,y:integer;
end;
var
ans:longint;
n,s:integer;
f:array[1..300,1..300] of longint;
w:array[1..300,0..300] of integer;
d:array[1..300] of longint;
a:array[0..300] of longint;
p:node;
procedure init;
var
i,j,p,q,v:integer;
begin
fillchar(f,sizeof(f),$5F);
readln(n,s);
for i:=1 to n do
begin
f[i,i]:=0;
for j:=i+1 to n do
begin
f[i,j]:=1000000;
f[j,i]:=1000000;
end;
end;
for i:=1 to n-1 do
begin
readln(p,q,v);
f[p,q]:=v;
f[q,p]:=v;
inc(w[p,0]);
w[p,w[p,0]]:=q;
inc(w[q,0]);
w[q,w[q,0]]:=p;
end;
end;
procedure floyd;
var
i,j,k:integer;
begin
for k:=1 to n do
for i:=1 to n do
for j:=1 to n do
if f[i,k]+f[k,j]<f[i,j] then f[i,j]:=f[i,k]+f[k,j];
end;
procedure findd;
var
i,j:integer;
max:longint;
begin
fillchar(p,sizeof(p),0);
max:=-maxint;
for i:=1 to n-1 do
for j:=i+1 to n do
if f[i,j]>max then
begin
p.x:=i;
p.y:=j;
max:=f[i,j];
end
else if f[i,j]=max then
begin
p.x:=i;
p.y:=j;
end;
end;
procedure findecc;
var
flag:array[1..300] of boolean;
i,j,k,l,r,pre:longint;
max:longint;
begin
ans:=maxint;
i:=1;
fillchar(a,sizeof(a),0);
fillchar(flag,sizeof(flag),false);
r:=p.x;
a[0]:=1;
a[1]:=r;
flag[r]:=true;
while r<>p.y do
begin
l:=r;
for j:=1 to w[l,0] do
if (f[p.x,w[l,j]]+f[w[l,j],p.y]=f[p.x,p.y]) and not flag[w[l,j]] then
begin
inc(a[0]);
a[a[0]]:=w[l,j];
r:=w[l,j];
flag[w[l,j]]:=true;
break;
end;
end;
fillchar(d,sizeof(d),$5F);
for j:=1 to a[0] do
begin
l:=a[j];
for k:=1 to n do
d[k]:=f[l,k];
for k:=j+1 to a[0] do
begin
r:=a[k];
if f[l,r]<=s then
begin
for pre:=1 to n do
if f[r,pre]<d[pre] then d[pre]:=f[r,pre];
end;
end;
max:=-maxint;
for k:=1 to n do
if d[k]>max then
begin
max:=d[k];
if max>=ans then break;
end;
if max<ans then ans:=max;
end;
writeln(ans);
end;
begin
assign(input,'core.in');
reset(input);
assign(output,'core.out');
rewrite(output);
init;
floyd;
findd;
findecc;
close(input);
close(output);
end.
