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 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
aaaajfaaaaOutput:
2
Notice: new testcases added
这题就是clj大牛在讲课中说到的题目，求多串的最长公共子串..
但是很多人用sa，nlogn的复杂度仍不能ac..
因为 spoj太慢了..
所以要用SAM来做，需要注意的一点是在能找到某个点可行时，要遍历所有的fail指针，因为这些也都可行..
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int Maxn = 100010;
int t, now, tot, F[Maxn*2], d[Maxn*2], ch[Maxn*2][26];
char s[Maxn]; int len;
int d2[Maxn*2], d1[Maxn*2];
int _min ( int x, int y ){ return x < y ? x : y; }
int _max ( int x, int y ){ return x > y ? x : y; }
int copy ( int p, int c ){int x = ++tot, y = ch[p][c];d[x] = d[p]+1;for ( int i = 0; i < 26; i ++ ) ch[x][i] = ch[y][i];F[x] = F[y]; F[y] = x;while ( ~p && ch[p][c] == y ){ ch[p][c] = x; p = F[p]; }return x;
}
void add ( int c ){int p, o;if ( p = ch[now][c] ){if ( d[p] != d[now]+1 ) copy ( now, c );now = ch[now][c];}else {d[o=++tot] = d[now]+1; p = now; now = o;while ( ~p && !ch[p][c] ){ ch[p][c] = o; p = F[p]; }F[o] = ~p ? ( d[ch[p][c]] == d[p]+1 ? ch[p][c] : copy ( p, c ) ) : 0;}
}
int main (){int i, j, k;F[0] = -1;scanf ( "%s", s+1 ); len = strlen (s+1);now = 0;for ( i = 1; i <= len; i ++ ) add (s[i]-'a');for ( i = 1; i <= tot; i ++ ) d2[i] = d[i];while ( scanf ( "%s", s+1 ) != EOF ){len = strlen (s+1);int o = 0; now = 0;for ( i = 1; i <= tot; i ++ ) d1[i] = 0;for ( i = 1; i <= len; i ++ ){int c = s[i]-'a';while ( ~o && !ch[o][c] ){ o = F[o]; now = o ? d[o] : 0; }if ( o < 0 ) o = now = 0; else o = ch[o][c], now ++;if ( now <= d1[o] ) continue; else d1[o] = now;int y = F[o]; while ( ~y && d1[y] < d[y] ) d1[y] = d[y], y = F[y];}for ( i = 1; i <= tot; i ++ ) d2[i] = _min ( d2[i], d1[i] );}int ans = 0;for ( i = 1; i <= tot; i ++ ) ans = _max ( ans, d2[i] );printf ( "%d\n", ans );return 0;
}