模板：
1.快排：
var
rank,height,h:array [0..300000] of qword;
//rank[i]:i的后缀的排名；
//height[i]:lcp(s[a[i,2]...len],s[a[i-1,2]...len])；
//h[i]:lcp(s[i...len],s[a[rank[i]-1,2]...len]);
//h[i]>=h[i]-1;
a:array [1..300000,1..2] of qword;
f:array [1..300000] of boolean;
s:string;
i,j,m,n,k,mm:longint;
x:qword;
procedure cha(l,r:longint);
var
i,j:longint;
mid,t:qword;
begin
i:=l;
j:=r;
mid:=a[(l+r) div 2,1];
while i<=j do
begin
while a[i,1]<mid do inc(i);
while a[j,1]>mid do dec(j);
if (i<=j) then
begin
t:=a[i,1];
a[i,1]:=a[j,1];
a[j,1]:=t;
t:=a[i,2];
a[i,2]:=a[j,2];
a[j,2]:=t;
inc(i);
dec(j);
end;
end;
if j>l then cha(l,j);
if i<r then cha(i,r);
end;
begin
assign(input,‘sss.in‘);
assign(output,‘sss.out‘);
reset(input);
rewrite(output);
readln(s);
for i:=1 to length(s) do
begin
a[i,1]:=ord(s[i])-ord(‘a‘)+1;
a[i,2]:=i;
rank[i]:=a[i,1];
end;
m:=1;
while n<>length(s) do
begin
n:=0;
if m<>0 then
for i:=1 to length(s) do
begin
x:=rank[i]*1000000;
x:=x+rank[i+m];
a[i,1]:=x;
a[i,2]:=i;
end;
cha(1,length(s));
n:=1;
rank[a[1,2]]:=1;
for i:=2 to length(s) do
begin
if a[i,1]<>a[i-1,1] then
inc(n);
rank[a[i,2]]:=n;
end;
m:=m*2;
end;
for i:=1 to length(s) do
write(a[i,2],‘ ‘);
writeln;
j:=0;
for i:=1 to length(s) do
begin
if j>0 then dec(j);
while (rank[i]-1<>0) and (s[i+j]=s[a[rank[i]-1,2]+j]) do inc(j);
h[i]:=j;
height[rank[i]]:=j;
end;
for i:=1 to length(s) do
write(height[i],‘ ‘);
writeln;
for i:=1 to length(s) do
write(h[i],‘ ‘);
writeln;
close(input);
close(output);
end.
2.桶排
var
rank,sa,lis,a:array [1..10000] of longint;
f:array [1..10000] of boolean;
s:string;
i,j,m,n,k:longint;
begin
readln(s);
m:=1;
for i:=1 to length(s) do
rank[i]:=ord(s[i])-ord(‘a‘)+1;
while n<>length(s) do
begin
n:=0;
for i:=1 to length(s) do
a[i]:=rank[i]*100+rank[i+m];
fillchar(f,sizeof(f),false);
for i:=1 to length(s) do
begin
if not f[a[i]] then
begin
inc(n);
f[a[i]]:=true;
end;
end;
k:=0;
for i:=1 to 2626 do
if f[i] then
begin
inc(k);
lis[i]:=k;
end;
for i:=1 to length(s) do
rank[i]:=lis[a[i]];
m:=m*2;
end;
for i:=1 to length(s) do
write(rank[i],‘ ‘);
writeln;
end.
倍增算法解释：
样例1：
sa:1 2 3 5 1 1 2 4
第一次排序rank:102 203 305 501 101 102 204 400
sa:2 3 5 7 1 2 4 6
第二次排序rank:205 307 501 702 104 206 400 600
sa:2 4 6 8 1 4 5 7
样例2：
sa:1 1 2 1 1 1 1 2
第一次排序rank:101 102 201 101 101 101 102 200
sa:1 2 4 1 1 1 2 3
第二次排序rank:104 201 401 101 102 103 200 300
sa:4 6 8 1 2 3 5 7
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