学习请看clj冬令营的讲稿吧，这主要是从SAM讲的
主要是从后缀树方面讲的，殊途同归
不过个人感觉从后缀树的方面更容易理解……
这里简单的说一下SAM的几个重要的性质
1.在SAM上，起点到任意一点的所有路径无重复的识别了所有子串
2.每个子串s出现的次数即ST(s)的right集合的大小（即子串s出现的所有右端点r的集合)
具体的就是parent树上子树right集合的并
3.（这是我认为最美妙的性质）SAM所构建出来的parent树与对原串逆序所购建出来的后缀树节点一一对应
具体的，SAM上每个节点最长接受子串=根节点到后缀树上每个点路径对应的字符串的逆序
4. 自动机的go[state][char]在parent树（后缀树）上的含义为（逆序）后缀树state节点代表的后缀前面加一个字符char的串所对应的节点
看图，（吐槽一下fhq博客的回文串）
一目了然了吧，正是有SAM在构造的过程中把一棵后缀树构造出来了，所以后缀数组能做的题目基本上SAM也能做
而且很多跟子串有关的题目，运用SAM明显比较简单
下面随便扯几个题目
bzoj3998 运用性质2，然后在拓扑图上维护一个前缀和即可
1 var mx,sa,fa:array[0..1000010] of longint;
2 sum,w:array[0..1000010] of int64;
3 go:array[0..1000010,'a'..'z'] of longint;
4 i,n,ty,k,x,j,t,last:longint;
5 s:ansistring;
6 c:char;
7
8 procedure add(c:char);
9 var p,np,nq,q:longint;
10 begin
11 inc(t); p:=last;
12 last:=t; np:=t; w[np]:=1;
13 mx[np]:=mx[p]+1;
14 while (p<>0) and (go[p,c]=0) do
15 begin
16 go[p,c]:=np;
17 p:=fa[p];
18 end;
19 if p=0 then fa[np]:=1
20 else begin
21 q:=go[p,c];
22 if mx[p]+1=mx[q] then fa[np]:=q
23 else begin
24 inc(t); nq:=t;
25 mx[nq]:=mx[p]+1;
26 go[nq]:=go[q];
27 fa[nq]:=fa[q];
28 fa[q]:=nq; fa[np]:=nq;
29 while go[p,c]=q do
30 begin
31 go[p,c]:=nq;
32 p:=fa[p];
33 end;
34 end;
35 end;
36 end;
37
38 procedure pre;
39 var c:char;
40 begin
41 for i:=1 to t do
42 inc(sum[mx[i]]);
43 for i:=1 to n do
44 inc(sum[i],sum[i-1]);
45 for i:=t downto 1 do
46 begin
47 sa[sum[mx[i]]]:=i;
48 dec(sum[mx[i]]);
49 end;
50 fillchar(sum,sizeof(sum),0);
51 for i:=t downto 1 do
52 begin
53 x:=sa[i];
54 if ty=1 then w[fa[x]]:=w[fa[x]]+w[x]
55 else w[x]:=1;
56 end;
57 w[1]:=0;
58 for i:=t downto 1 do
59 begin
60 x:=sa[i];
61 sum[x]:=w[x];
62 for c:='a' to 'z' do
63 sum[x]:=sum[x]+sum[go[x,c]];
64 end;
65 end;
66
67 procedure get(x,k:longint);
68 var c:char;
69 y:longint;
70 begin
71 while true do
72 begin
73 if k<=w[x] then exit;
74 k:=k-w[x];
75 for c:='a' to 'z' do
76 begin
77 y:=go[x,c];
78 if y<>0 then
79 begin
80 if k<=sum[y] then
81 begin
82 write(c);
83 x:=y;
84 break;
85 end;
86 k:=k-sum[y];
87 end;
88 end;
89 end;
90 end;
91
92 begin
93 readln(s);
94 n:=length(s);
95 last:=1; t:=1;
96 for i:=1 to n do
97 add(s[i]);
98 readln(ty,k);
99 pre;
100 if k>sum[1] then writeln(-1)
101 else get(1,k);
102 end.
3998
bzoj3676 先用manacher找出所有本质不同回文串，然后放到SAM上跑即可
怎么跑？当然不是一个个匹配啦，把一个子串当作某个后缀的前缀，然后找到后缀对应的节点
然后倍增往上提~
（听说有个东西叫回文自动机？……）
1 var go:array[0..600010,'a'..'z'] of longint;
2 sa,sum,d,fa,mx:array[0..600010] of longint;
3 p,loc:array[0..310000] of longint;
4 anc:array[0..600010,0..18] of longint;
5 w:array[0..600010] of int64;
6 s:array[0..310000] of char;
7 last,t,n:longint;
8 ans:int64;
9
10 function min(a,b:longint):longint;
11 begin
12 if a>b then exit(b) else exit(a);
13 end;
14
15 function max(a,b:int64):int64;
16 begin
17 if a>b then exit(a) else exit(b);
18 end;
19
20 procedure add(c:char);
21 var p,q,np,nq:longint;
22 begin
23 p:=last;
24 inc(t); last:=t; np:=t;
25 w[np]:=1; mx[np]:=mx[p]+1;
26 while (p<>0) and (go[p,c]=0) do
27 begin
28 go[p,c]:=np;
29 p:=fa[p];
30 end;
31 if p=0 then fa[np]:=1
32 else begin
33 q:=go[p,c];
34 if mx[q]=mx[p]+1 then fa[np]:=q
35 else begin
36 inc(t); nq:=t;
37 mx[nq]:=mx[p]+1;
38 go[nq]:=go[q];
39 fa[nq]:=fa[q];
40 fa[q]:=nq; fa[np]:=nq;
41 while go[p,c]=q do
42 begin
43 go[p,c]:=nq;
44 p:=fa[p];
45 end;
46 end;
47 end;
48 end;
49
50 procedure prework;
51 var x,y,i,j:longint; c:char;
52 begin
53 for i:=1 to t do
54 inc(sum[mx[i]]);
55 for i:=1 to n do
56 inc(sum[i],sum[i-1]);
57 for i:=t downto 1 do
58 begin
59 sa[sum[mx[i]]]:=i;
60 dec(sum[mx[i]]);
61 end;
62 for i:=t downto 1 do
63 begin
64 x:=sa[i];
65 w[fa[x]]:=w[fa[x]]+w[x];
66 end;
67 for i:=1 to t do
68 begin
69 x:=sa[i];
70 d[x]:=d[fa[x]]+1;
71 anc[x,0]:=fa[x];
72 for j:=1 to 18 do
73 begin
74 y:=anc[x,j-1];
75 if anc[y,j-1]<>0 then anc[x,j]:=anc[y,j-1] else break;
76 end;
77 end;
78 end;
79
80 procedure getchar;
81 var ch:char;
82 begin
83 read(ch);
84 while (ch>='a') and (ch<='z') do
85 begin
86 inc(n);
87 s[n]:=ch;
88 add(ch);
89 loc[n]:=last;
90 read(ch);
91 end;
92 end;
93
94 procedure find(l,r:longint);
95 var x,i:longint;
96 begin
97 x:=loc[r];
98 for i:=18 downto 0 do
99 if mx[anc[x,i]]>=r-l+1 then x:=anc[x,i];
100 ans:=max(ans,int64(r-l+1)*w[x]);
101 end;
102
103 procedure manacher;
104 var i,k,right:longint;
105 begin
106 right:=0; k:=0;
107 for i:=1 to n do
108 begin
109 if right>i then p[i]:=min(right-i,p[2*k-i])
110 else begin
111 p[i]:=1;
112 find(i-p[i]+1,i+p[i]-1);
113 end;
114 while s[i+p[i]]=s[i-p[i]] do
115 begin
116 inc(p[i]);
117 find(i-p[i]+1,i+p[i]-1);
118 end;
119 if p[i]+i>right then
120 begin
121 k:=i;
122 right:=p[i]+i;
123 end;
124 end;
125 right:=0; k:=0;
126 for i:=1 to n do
127 begin
128 if right>i then p[i]:=min(right-i,p[2*k-i-1])
129 else p[i]:=0;
130 while s[i+p[i]+1]=s[i-p[i]] do
131 begin
132 inc(p[i]);
133 find(i-p[i]+1,i+p[i]);
134 end;
135 if p[i]+i>right then
136 begin
137 k:=i;
138 right:=p[i]+i;
139 end;
140 end;
141 end;
142
143 begin
144 last:=1; t:=1;
145 getchar;
146 s[0]:='#'; s[n+1]:='$';
147 prework;
148 manacher;
149 writeln(ans);
150 end.
3676
bzoj1396 2865（注意2865数据规模大了，SAM注意卡空间）
注意出现一次的子串s ST(s)一定是parent树上的叶子节点
clj ppt中有一个性质，一个节点p可接受的子串长度=[min(s),max(s)]=[max(fa[p])+1,max(s)]
所以我们反序构造SAM，叶子节点代表了一个后缀i
这样我们就可以更新，对于[i,i+max(fa[s])+1] 我们用max(fa[s])+1更新
对于j属于[i+max(fa[s])+2,n]我们用j-i更新
相当于一个线段覆盖问题，只不过一种线段斜率为0，一种线段斜率为1，我们可以分别用线段树维护
1 const inf=100000007;
2 var go:array[0..1000010,'a'..'z'] of longint;
3 mx,fa,w:array[0..1000010] of longint;
4 v:array[0..1000010] of boolean;
5 tree:array[0..500010*3,0..1] of longint;
6 i,n,t,last:longint;
7 s:ansistring;
8
9 function min(a,b:longint):longint;
10 begin
11 if a>b then exit(b) else exit(a);
12 end;
13
14 procedure add(c:char);
15 var p,q,np,nq:longint;
16 begin
17 p:=last;
18 inc(t); last:=t; np:=t;
19 mx[np]:=mx[p]+1;
20 while (p<>0) and (go[p,c]=0) do
21 begin
22 go[p,c]:=np;
23 p:=fa[p];
24 end;
25 if p=0 then fa[np]:=1
26 else begin
27 q:=go[p,c];
28 if mx[q]=mx[p]+1 then fa[np]:=q
29 else begin
30 inc(t); nq:=t;
31 mx[nq]:=mx[p]+1;
32 go[nq]:=go[q];
33 fa[nq]:=fa[q];
34 fa[q]:=nq; fa[np]:=nq;
35 while go[p,c]=q do
36 begin
37 go[p,c]:=nq;
38 p:=fa[p];
39 end;
40 end;
41 end;
42 end;
43
44 procedure build(i,l,r:longint);
45 var m:longint;
46 begin
47 tree[i,0]:=inf;
48 tree[i,1]:=inf;
49 if l<>r then
50 begin
51 m:=(l+r) shr 1;
52 build(i*2,l,m);
53 build(i*2+1,m+1,r);
54 end;
55 end;
56
57 procedure cov(i,q,z:longint);
58 begin
59 tree[i,q]:=min(tree[i,q],z);
60 end;
61
62 procedure push(i,q:longint);
63 begin
64 cov(i*2,q,tree[i,q]);
65 cov(i*2+1,q,tree[i,q]);
66 tree[i,q]:=inf;
67 end;
68
69 procedure work(i,l,r,q,x,y,z:longint);
70 var m:longint;
71 begin
72 if (x<=l) and (y>=r) then cov(i,q,z)
73 else begin
74 if tree[i,q]<inf then push(i,q);
75 m:=(l+r) shr 1;
76 if x<=m then work(i*2,l,m,q,x,y,z);
77 if y>m then work(i*2+1,m+1,r,q,x,y,z);
78 end;
79 end;
80
81 procedure ask(i,l,r:longint);
82 var m:longint;
83 begin
84 if l=r then writeln(min(n,min(tree[i,0],tree[i,1]+l)))
85 else begin
86 if tree[i,0]<inf then push(i,0);
87 if tree[i,1]<inf then push(i,1);
88 m:=(l+r) shr 1;
89 ask(i*2,l,m);
90 ask(i*2+1,m+1,r);
91 end;
92 end;
93
94 begin
95 readln(s);
96 n:=length(s);
97 t:=1; last:=1;
98 for i:=n downto 1 do
99 begin
100 add(s[i]);
101 w[last]:=i;
102 end;
103 for i:=1 to t do
104 v[fa[i]]:=true;
105
106 build(1,1,n);
107 for i:=1 to t do
108 if not v[i] then
109 begin
110 work(1,1,n,0,w[i],w[i]+mx[fa[i]],mx[fa[i]]+1);
111 if w[i]+mx[fa[i]]<n then
112 work(1,1,n,1,w[i]+mx[fa[i]]+1,n,-w[i]+1);
113 end;
114 ask(1,1,n);
115 end.
2865
bzoj2946 如果用SA做，是经典的二分然后对height分组
用SAM怎么做呢？我们可以先把一个串的SAM建出来，然后用其它串匹配
得出每个节点在这个串匹配的最大长度，然后可以得到每个节点在所有串匹配的最大值的最小值
然后取所有节点这个值的最大值即可（很拗口，但是很明显）
1 var go:array[0..20010,'a'..'z'] of longint;
2 sum,mx,fa,f,a,sa:array[0..20010] of longint;
3 ans,m,last,t,i,n,l,j,p,len:longint;
4 s:ansistring;
5
6 function min(a,b:longint):longint;
7 begin
8 if a>b then exit(b) else exit(a);
9 end;
10
11 function max(a,b:longint):longint;
12 begin
13 if a>b then exit(a) else exit(b);
14 end;
15
16 procedure add(c:char);
17 var p,q,np,nq:longint;
18 begin
19 p:=last;
20 inc(t); last:=t; np:=t;
21 mx[np]:=mx[p]+1;
22 while (p<>0) and (go[p,c]=0) do
23 begin
24 go[p,c]:=np;
25 p:=fa[p];
26 end;
27 if p=0 then fa[np]:=1
28 else begin
29 q:=go[p,c];
30 if mx[q]=mx[p]+1 then fa[np]:=q
31 else begin
32 inc(t); nq:=t;
33 mx[nq]:=mx[p]+1;
34 go[nq]:=go[q];
35 fa[nq]:=fa[q];
36 fa[q]:=nq; fa[np]:=nq;
37 while go[p,c]=q do
38 begin
39 go[p,c]:=nq;
40 p:=fa[p];
41 end;
42 end;
43 end;
44 end;
45
46 procedure pre;
47 var i:longint;
48 begin
49 for i:=1 to t do
50 begin
51 inc(sum[mx[i]]);
52 f[i]:=mx[i];
53 end;
54 for i:=1 to n do
55 inc(sum[i],sum[i-1]);
56 for i:=t downto 1 do
57 begin
58 sa[sum[mx[i]]]:=i;
59 dec(sum[mx[i]]);
60 end;
61 end;
62
63 begin
64 readln(m);
65 readln(s);
66 n:=length(s);
67 last:=1; t:=1;
68 for i:=1 to n do
69 add(s[i]);
70 pre;
71 for i:=2 to m do
72 begin
73 readln(s);
74 l:=length(s);
75 p:=1; len:=0;
76 fillchar(a,sizeof(a),0);
77 for j:=1 to l do
78 begin
79 while (p<>0) and (go[p,s[j]]=0) do p:=fa[p];
80 if p=0 then
81 begin
82 p:=1;
83 len:=0;
84 end
85 else begin
86 len:=min(len,mx[p])+1;
87 p:=go[p,s[j]];
88 end;
89 a[p]:=max(a[p],len);
90 end;
91 for j:=t downto 1 do
92 begin
93 p:=sa[j];
94 a[fa[p]]:=max(a[fa[p]],a[p]);
95 end;
96 for j:=1 to t do
97 f[j]:=min(f[j],a[j]);
98 end;
99 for i:=1 to t do
100 ans:=max(ans,f[i]);
101 writeln(ans);
102 end.
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