地址：http://www.spoj.com/problems/LCS2/
题面：
LCS2 - Longest Common Substring II
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A string is finite sequence of characters over a non-empty finite set Σ.
In this problem, Σ is the set of lowercase letters.
Substring, also called factor, is a consecutive sequence of characters occurrences at least once in a string.
Now your task is a bit harder, for some given strings, find the length of the longest common substring of them.
Here common substring means a substring of two or more strings.
Input
The input contains at most 10 lines, each line consists of no more than 100000 lowercase letters, representing a string.
Output
The length of the longest common substring. If such string doesn‘t exist, print "0" instead.
Example
Input: alsdfkjfjkdsal fdjskalajfkdsla aaaajfaaaa Output: 2
Notice: new testcases added
思路：
对第一个穿构建sam，然后让后面的串在sam上走，同时记录每个串跑完后各个状态的所能匹配的最长长度，然后对所有串取个最小值，之后再取个最大值。
注意：如果状态p能够到达，那么状态fa[p]也必然能够到达，但可能在匹配时不会走到fa[p]状态，所以当p可达时，要更新fa[p]。
1 #include <bits/stdc++.h>
2
3 using namespace std;
4
5 struct SAM
6 {
7 static const int MAXN = 100001<<1;//大小为字符串长度两倍
8 static const int LetterSize = 26;
9
10 int tot, last, ch[MAXN][LetterSize], fa[MAXN], len[MAXN];
11 int sum[MAXN], tp[MAXN]; //用于拓扑排序，tp为排序后的数组
12
13 void init( void)
14 {
15 last = tot = 1;
16 len[1] = 0;
17 memset(ch,0,sizeof ch);
18 memset(fa,0,sizeof fa);
19 }
20
21 void add( int x)
22 {
23 int p = last, np = last = ++tot;
24 len[np] = len[p] + 1;
25 while( p && !ch[p][x]) ch[p][x] = np, p = fa[p];
26 if( p == 0)
27 fa[np] = 1;
28 else
29 {
30 int q = ch[p][x];
31 if( len[q] == len[p] + 1)
32 fa[np] = q;
33 else
34 {
35 int nq = ++tot;
36 memcpy( ch[nq], ch[q], sizeof ch[q]);
37 len[nq] = len[p] + 1, fa[nq] = fa[q], fa[q] = fa[np] = nq;
38 while( p && ch[p][x] == q) ch[p][x] = nq, p = fa[p];
39 }
40 }
41 }
42
43 void toposort( void)
44 {
45 for(int i = 1; i <= len[last]; i++) sum[i] = 0;
46 for(int i = 1; i <= tot; i++) sum[len[i]]++;
47 for(int i = 1; i <= len[last]; i++) sum[i] += sum[i-1];
48 for(int i = 1; i <= tot; i++) tp[sum[len[i]]--] = i;
49 }
50 } sam;
51 char ss[100004];
52 int ans,mx[100001<<1],mi[100001<<1];
53
54 int main(void)
55 {
56 //freopen("in.acm","r",stdin);
57 sam.init();
58 scanf("%s",ss);
59 for(int i=0,len=strlen(ss);i<len;i++) sam.add(ss[i]-‘a‘);
60 for(int i=1;i<=sam.tot;i++) mi[i]=1e6;
61 sam.toposort();
62 while(~scanf("%s",ss))
63 {
64 for(int i=0,len=strlen(ss),p=1,cnt=0;i<len;i++)
65 {
66 int c=ss[i]-‘a‘;
67 if(sam.ch[p][c])
68 p=sam.ch[p][c],cnt++;
69 else
70 {
71 while(p&&!sam.ch[p][c]) p=sam.fa[p];
72 if(!p)
73 p=1,cnt=0;
74 else
75 cnt=sam.len[p]+1,p=sam.ch[p][c];
76 }
77 mx[p]=max(mx[p],cnt);
78 }
79 for(int i=sam.tot;i;i--)
80 {
81 int p=sam.tp[i];
82 mi[p]=min(mx[p],mi[p]);
83 if(mx[p]&&sam.fa[p]) mx[sam.fa[p]]=sam.len[sam.fa[p]];
84 mx[p]=0;
85 }
86 }
87 for(int i=1;i<=sam.tot;i++) ans=max(ans,mi[i]);
88 printf("%d\n",ans);
89 return 0;
90 }