后缀数组实现

先贴这里,免得忘了

program sa;type int=longint;var        i,j,k,m,n,l,max:int;        a,b,c,d,x,r:array[0..1000000]of int;        ch:char;procedure main;begin        l:=1;        repeat                for i:=0 to max do begin b[i]:=0;c[i]:=0;end;                for i:=1 to n do begin inc(b[a[i]]);inc(c[a[i+l]]);end;                for i:=1 to max do begin inc(b[i],b[i-1]);inc(c[i],c[i-1]);end;                for i:=1 to n do begin d[c[a[i+l]]]:=i;dec(c[a[i+l]]);end;                for i:=n downto 1 do begin k:=d[i];r[b[a[k]]]:=d[i];dec(b[a[k]]);end;                max:=0;                for i:=1 to n do begin                        if(a[r[i-1]]<>a[r[i]])or(a[r[i-1]+l]<>a[r[i]+l])then inc(max);                        x[r[i]]:=max;                end;                for i:=1 to n do a[i]:=x[i];                l:=l<<1;        until max=n;end;begin        assign(input,'sa.in');reset(input);        assign(output,'sa.out');rewrite(output);        while not eoln do begin                inc(n);read(ch);a[n]:=ord(ch);                if a[n]>max then max:=a[n];        end;        main;        for i:=1 to n do writeln(a[i]);        close(input);close(output);end.

如果使用指针实现X,A的功能就可以少一个循环,应该会更快一点

最长公共子串:

program sa;type int=longint;var        l,i,j,k,m,n,max,ans:int;        a,b,c,d,r,x,h,st:array[0..500000]of int;        ch:char;procedure main;begin        l:=1;max:=200;        repeat                for i:=0 to max do begin c[i]:=0;b[i]:=0;end;                for i:=1 to n do begin inc(b[a[i]]);inc(c[a[i+l]]);end;                for i:=1 to max do begin inc(b[i],b[i-1]);inc(c[i],c[i-1]);end;                for i:=1 to n do begin d[c[a[i+l]]]:=i;dec(c[a[i+l]]);end;                for i:=n downto 1 do begin r[b[a[d[i]]]]:=d[i];dec(b[a[d[i]]]);end;                max:=0;                for i:=1 to n do begin                        if(a[r[i]]<>a[r[i-1]])or(a[r[i]+l]<>a[r[i-1]+l])then inc(max);                        x[r[i]]:=max;                end;                for i:=1 to n do a[i]:=x[i];                l:=l<<1;        until max=n;        for i:=1 to n do b[a[i]]:=i;        for i:=1 to n do begin                if h[i-1]>0 then h[i]:=h[i-1]-1;                while st[b[a[i]-1]+h[i]]=st[i+h[i]] do inc(h[i]);        end;        for i:=1 to n do x[a[i]]:=h[i];        for i:=2 to n do if((b[i]<m)xor(b[i-1]<m))and(ans<x[i])                then ans:=x[i];end;//a->rank,b->sa,h->h,x->heightbegin        assign(input,'data.txt');reset(input);        assign(output,'ans.txt');rewrite(output);        while not eoln do begin read(ch);inc(n);st[n]:=ord(ch);end;        m:=n+1;readln;        inc(n);st[n]:=200;        while not eoln do begin read(ch);inc(n);st[n]:=ord(ch);end;        a:=st;main;        write(ans);        close(input);close(output);end.