我的方法是设f[i]为长度为i的种数，预处理时考虑一进一出的路径

用g表示s出发，t结束，中间长度为l的路径是否算过了，避免重复

if j=maxn do

begin

c:=c[i] div maxn+c;

c[i]:=c[i] mod maxn;

end;

while (c[0]>1)and(c[c[0]]=0) do dec(c[0]);

exit(c);

end;

procedure print(a:xtype);

var

i,j:longint;

begin

if a[0]=0 then begin write('0'); halt; end;

write(a[a[0]]);

for i:=a[0]-1 downto 1 do

begin

if a[i]>=10000000 then else

if a[i]>=1000000 then write('0') else

if a[i]>=100000 then write('00') else

if a[i]>=10000 then write('000') else

if a[i]>=1000 then write('0000') else

if a[i]>=100 then write('00000') else

if a[i]>=10 then write('000000') else

write('0000000');

write(a[i]);

end;

end;

begin

read(n,l);

f[0][0]:=1; f[0][1]:=1;

for i:=1 to n do read(next[i]);

for i:=1 to n do read(e[i]);

for i:=1 to n do

begin

s:=i; t:=i; tot:=e[i];

if tot+e[t]n then t:=1;

until tot+e[t]>l;

end;

for i:=1 to l do

for j:=1 to i-1 do

if j