SPOJ LCSLongest Common Substring

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.

Example

Input:alsdfkjfjkdsalfdjskalajfkdslaOutput:3

Notice: new testcases added

后缀自动机

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;const int maxn = 300005;char s[maxn];class SAM{const static int maxn = 500005;   //节点个数  const static int size = 26;     //字符的范围  const static char base = 'a';     //字符的基准  class node{public:node *fa, *next[size];int len, cnt;node* clear(int x){fa = 0; len = x;cnt = 0;memset(next, 0, sizeof(next));return this;}}nd[maxn];                      //节点的设置  node *root, *last;              //根节点,上一个节点  int tot;                        //总节点数  public:void clear(){last = root = &nd[tot = 0];nd[0].clear(0);}                               //初始化  void insert(char ch){node *p = last, *np = nd[++tot].clear(p->len + 1);last = np;int x = ch - base;while (p&&p->next[x] == 0) p->next[x] = np, p = p->fa;if (p == 0) { np->fa = root; return; }node* q = p->next[x];if (p->len + 1 == q->len) { np->fa = q; return; }node *nq = nd[++tot].clear(p->len + 1);for (int i = 0; i < size; i++)if (q->next[i]) nq->next[i] = q->next[i];nq->fa = q->fa;q->fa = np->fa = nq;while (p &&p->next[x] == q) p->next[x] = nq, p = p->fa;}                               //插入操作  void find(char *ch){node *s = root;int maxlen = 0, ans = 0;for (int i = 0; ch[i]; i++){int x = ch[i] - base;if (s->next[x]) {s = s->next[x];s->cnt = max(s->cnt, ++maxlen);}else{while (s&&!s->next[x]) s = s->fa;if (s == 0) s = root,maxlen = 0;else  maxlen = s->len + 1, s = s->next[x];}ans = max(ans, maxlen);}printf("%d\n", ans);}}sam;int main(){while (~scanf("%s", s)){sam.clear();for (int i = 0; s[i]; i++) sam.insert(s[i]);scanf("%s", s);sam.find(s);}return 0;}