题目连接：

https://vjudge.net/problem/SPOJ-LCS2

Description

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 simple, for two 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 exactly two lines, each line consists of no more than 250000 lowercase letters, representing a string.

Output

The length of the longest common substring. If such string doesn't exist, print "0" instead.

Sample Input

alsdfkjfjkdsal
fdjskalajfkdsla

Sample Output

3

Hint

题意

求两个串的最长公共子串长度

题解：

对a串建后缀自动机，b串在建好的自动机上跑，跑的过程和建树是一样的，如果Next[p][c]存在则nowlen++，否则p回退到fa[p]直到Next[p][c]存在，并更新nowlen为len[p]+1,对整个过程的nowlen取max就是答案

代码

#include <cstdio>
#include <cstring>
#include <algorithm>using namespace std;
const int mx = 1e6+5;struct SAM_automaton {int Next[mx][26], len[mx], fa[mx];int last, tot;int newnode() {tot++;for (int i = 0; i < 26; i++) Next[tot][i] = 0;return tot;}void init() {tot = 0;last = newnode();}void add(int c) {int p = last;int np = last = newnode();len[np] = len[p] + 1;while (p && !Next[p][c]) {Next[p][c] = np;p = fa[p];}if (!p) fa[np] = 1;else {int q = Next[p][c];if (len[q] == len[p] + 1) fa[np] = q;else {int nq = newnode();len[nq] = len[p] + 1;fa[nq] = fa[q];for (int i = 0; i < 26; i++) Next[nq][i] = Next[q][i];fa[q] = fa[np] = nq;while (p && Next[p][c] == q) {Next[p][c] = nq;p = fa[p];}}}}
}SAM;char a[mx], b[mx], str[mx];int main() {SAM.init();scanf("%s%s", a, b);int lena = strlen(a);int lenb = strlen(b);for (int i = 0; i < lena; i++) SAM.add(a[i]-'a');int ans = 0, len = 0;for (int i = 0, p = 1; i < lenb; i++) {int c = b[i] - 'a';if (SAM.Next[p][c]) {len++;p = SAM.Next[p][c];} else {while (p && !SAM.Next[p][c]) p = SAM.fa[p];if (!p) {p = 1;len =0;} else {len = SAM.len[p] + 1;p = SAM.Next[p][c];}}ans = max(ans, len);}printf("%d\n", ans);    return 0;
}