1509 -- Glass Beads POJ

题意：求一个字符串的最小表示的开始下标

就当模板题写了

把字符串重复一遍，再建后缀自动机，贪心的选最小字典序在上面走len步

因为走出来的一定是子串，长度又是len，所以一定是原来的字符串旋转得到的，就解决了

 

 1 const
 2     maxn=50010;
 3 type
 4     node=record
 5       go:array[0..25]of longint;
 6       step,fa:longint;
 7     end;
 8 
 9 var
10     sam:array[0..maxn]of node;
11     s,ans:ansistring;
12     len,tot,last,t:longint;
13 
14 procedure add(x:longint);
15 var
16     now,new,q:longint;
17 begin
18     inc(tot);
19     now:=tot;
20     fillchar(sam[now].go,sizeof(sam[now].go),0);
21     sam[now].step:=sam[last].step+1;
22     while (last<>0)and(sam[last].go[x]=0) do
23       begin
24         sam[last].go[x]:=now;
25         last:=sam[last].fa;
26       end;
27     if last=0 then sam[now].fa:=1
28     else
29       begin
30         q:=sam[last].go[x];
31         if sam[q].step=sam[last].step+1 then sam[now].fa:=q
32         else
33           begin
34             inc(tot);
35             new:=tot;
36             sam[new]:=sam[q];
37             sam[q].fa:=new;
38             sam[now].fa:=new;
39             sam[new].step:=sam[last].step+1;
40             while (last<>0)and(sam[last].go[x]=q) do
41               begin
42                 sam[last].go[x]:=new;
43                 last:=sam[last].fa;
44               end;
45           end;
46       end;
47     last:=now;
48 end;
49 
50 procedure main;
51 var
52     i,j,k:longint;
53 begin
54     readln(s);
55     s:=s+s;
56     len:=length(s);
57     last:=1;
58     tot:=1;
59     sam[1].fa:=0;
60     sam[1].step:=0;
61     fillchar(sam[1].go,sizeof(sam[1].go),0);
62     for i:=1 to len do
63       add(ord(s[i])-ord('a'));
64     i:=1;
65     j:=0;
66     ans:='';
67     while j<len>>1 do
68       begin
69         inc(j);
70         for k:=0 to 25 do
71           if sam[i].go[k]<>0 then break;
72         ans:=ans+chr(k+ord('a'));
73         i:=sam[i].go[k];
74       end;
75     writeln(pos(ans,s));
76 end;
77 
78 begin
79     readln(t);
80     while t<>0 do
81       begin
82         main;
83         dec(t);
84       end;
85 end.

View Code

 

转载于:https://www.cnblogs.com/Randolph87/p/3617727.html