标签:
...其实暑假的时候写过一次,不过那时候对这道题理解不是很深,所以重写了一遍....
尝试用新的模版去写,然后发现新的模版里面我把fail并到next,以省去多次的while取点,但是对于这道题,fail是必须用到的,因为要DP...所以不能并进去...于是只能乖乖滚回去写原来的方法,每次都去往回while fail节点...然后就是pd数组的传递,要和fail所连的点保持一直,因为是BFS下来的...
然后...代码如下
1 const maxn=6419; 2 maxs=25; 3 vv=10007; 4 var 5 next:array[0..maxn,0..maxs] of longint; 6 pd:array[0..maxn] of boolean; 7 f:array[0..maxn] of longint; 8 q:array[0..maxn] of longint; 9 dp:array[0..maxn,0..105] of longint; 10 s:string; 11 head,tail,root,n,m,tot:longint; 12 13 procedure push(x:longint); begin inc(tail); q[tail]:=x; end; 14 15 function new:longint; 16 var i:longint; 17 begin 18 pd[tot]:=false; 19 for i:= 0 to maxs do next[tot,i]:=-1; 20 inc(tot); exit(tot-1); 21 end; 22 23 procedure insert(s:string); 24 var c,i,v:longint; 25 begin 26 v:=root; 27 for i:= 1 to length(s) do 28 begin 29 c:=ord(s[i])-65; 30 if next[v,c]=-1 then next[v,c]:=new; 31 v:=next[v,c]; 32 end; 33 pd[v]:=true; 34 end; 35 36 procedure build; 37 var v,tmp,i:longint; 38 begin 39 f[root]:=root; 40 head:=1; tail:=0; 41 for i:= 0 to maxs do 42 if next[root,i]<>-1 then 43 begin f[next[root,i]]:=root; push(next[root,i]); end; 44 while head<=tail do 45 begin 46 v:=q[head]; inc(head); 47 for i:= 0 to maxs do 48 if next[v,i]<>-1 then 49 begin 50 push(next[v,i]); 51 tmp:=f[v]; 52 while (tmp<>root) and (next[tmp,i]=-1) do tmp:=f[tmp]; 53 if next[tmp,i]<>-1 then tmp:=next[tmp,i]; 54 if tmp=-1 then f[next[v,i]]:=root else f[next[v,i]]:=tmp; 55 if not pd[next[v,i]] then pd[next[v,i]]:=pd[tmp]; 56 end; 57 end; 58 end; 59 60 procedure solve; 61 var i,j,k,v,cnt,ans:longint; 62 begin 63 dp[0,0]:=1; 64 for i:= 1 to m do 65 for j:= root to tot do 66 if (not pd[j]) and (dp[j,i-1]>0) then 67 for k:= 0 to maxs do 68 begin 69 v:=j; 70 while (next[v,k]=-1) and (v<>root) do v:=f[v]; 71 if next[v,k]<>-1 then v:=next[v,k]; 72 if not pd[v] then dp[v,i]:=(dp[v,i]+dp[j,i-1]) mod vv; 73 end; 74 cnt:=0; 75 for i:= 0 to tot do cnt:=(cnt+dp[i,m]) mod vv; 76 ans:=1; 77 for i:= 1 to m do ans:=(ans*26) mod vv; 78 ans:=(((ans-cnt) mod vv+vv) mod vv); 79 writeln(ans); 80 end; 81 82 procedure init; 83 var i:longint; 84 begin 85 readln(n,m); 86 root:=new; 87 for i:= 1 to n do 88 begin 89 readln(s); 90 insert(s); 91 end; 92 build; 93 end; 94 95 Begin 96 init; 97 solve 98 End.
标签:
原文地址:http://www.cnblogs.com/EC-Ecstasy/p/4183218.html