题目的意思是叶子不超过20个……听说当初zjoi不少人被坑
分别对每个叶子以它为根dfs出20个dfs树，这样整个树的任何一个子串，都是某个dfs树上一个点到它的一个子孙的路径
每个dfs树，根到叶子相当于一个串，这样相当于统计20*19个串的不同字串数目
这样我们很容易想到建立多串SAM（广义后缀树）统计不同的子串数即可
1 type node=record 2 po,next:longint; 3 end; 4 5 var go:array[0..4000010,0..10] of longint; 6 fa,mx:array[0..4000010] of longint; 7 c,d,p:array[0..100010] of longint; 8 e:array[0..200010] of node; 9 v:array[0..100010] of boolean; 10 t,i,len,x,y,n,m:longint; 11 ans:int64; 12 13 procedure add(x,y:longint); 14 begin 15 inc(len); 16 e[len].po:=y; 17 e[len].next:=p[x]; 18 p[x]:=len; 19 inc(d[y]); 20 end; 21 22 function work(c,last:longint):longint; 23 var np,q,p,nq:longint; 24 begin 25 p:=last; 26 if go[p,c]=0 then 27 begin 28 inc(t); work:=t; np:=t; 29 mx[np]:=mx[p]+1; 30 while (p>0) and (go[p,c]=0) do 31 begin 32 go[p,c]:=np; 33 p:=fa[p]; 34 end; 35 if p=0 then fa[np]:=1 36 else begin 37 q:=go[p,c]; 38 if mx[q]=mx[p]+1 then fa[np]:=q 39 else begin 40 inc(t); nq:=t; 41 mx[nq]:=mx[p]+1; 42 go[nq]:=go[q]; 43 fa[nq]:=fa[q]; 44 fa[q]:=nq; fa[np]:=nq; 45 while go[p,c]=q do 46 begin 47 go[p,c]:=nq; 48 p:=fa[p]; 49 end; 50 end; 51 end; 52 end 53 else begin 54 q:=go[p,c]; 55 if mx[q]=mx[p]+1 then exit(q) 56 else begin 57 inc(t); nq:=t; 58 mx[nq]:=mx[p]+1; 59 go[nq]:=go[q]; 60 fa[nq]:=fa[q]; 61 fa[q]:=nq; 62 while go[p,c]=q do 63 begin 64 go[p,c]:=nq; 65 p:=fa[p]; 66 end; 67 end; 68 exit(go[last,c]); 69 end; 70 end; 71 72 procedure dfs(x,last:longint); 73 var i,y:longint; 74 begin 75 last:=work(c[x],last); 76 v[x]:=true; 77 i:=p[x]; 78 while i<>0 do 79 begin 80 y:=e[i].po; 81 if not v[y] then dfs(y,last); 82 i:=e[i].next; 83 end; 84 end; 85 86 begin 87 readln(n,m); 88 for i:=1 to n do 89 read(c[i]); 90 for i:=1 to n-1 do 91 begin 92 readln(x,y); 93 add(x,y); 94 add(y,x); 95 end; 96 t:=1; 97 for i:=1 to n do 98 if d[i]=1 then 99 begin 100 fillchar(v,sizeof(v),false); 101 dfs(i,1); 102 end; 103 for i:=2 to t do 104 ans:=ans+mx[i]-mx[fa[i]]; 105 writeln(ans); 106 end.
