SPOJ LCS2 Longest Common Substring II
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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;}
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