SPOJ1811 LCS - Longest Common Substring(後綴自動機)

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:
alsdfkjfjkdsal
fdjskalajfkdsla

Output:
3

Notice: new testcases added

 

爲何都說這是後綴自動機的裸題啊。難道就我看不出來麼qwq、、

正解其實比較好想，先把第一個串扔到SAM裏面

對於第二個串依次枚舉，若是能匹配就匹配，不然就沿着parent邊走(怎麼這麼像AC自動機？？)，順便記錄一下最長長度

 

 

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 2 * 300000 + 1;
char s1[MAXN], s2[MAXN];
int N1, N2, tot = 1, root = 1, last = 1, len[MAXN], ch[MAXN][27], fa[MAXN];
void insert(int x) {
    int now = ++tot, pre = last; last = now; len[now] = len[pre] + 1;
    for(; pre && !ch[pre][x]; pre = fa[pre]) ch[pre][x] = now;
    if(!pre) fa[now] = root;
    else {
        int q = ch[pre][x];
        if(len[q] == len[pre] + 1) fa[now] = q;
        else {
            int nows = ++tot; 
            memcpy(ch[nows], ch[q], sizeof(ch[q]));
            len[nows] = len[pre] + 1;
            fa[nows] = fa[q]; fa[now] = fa[q] = nows;
            for(; pre && ch[pre][x] == q; pre = fa[pre]) ch[pre][x] = nows;                
        }
    } 

}
int main() {
#ifdef WIN32
    freopen("a.in", "r", stdin);
#endif
    scanf("%s %s", s1 + 1, s2 + 1);
    N1 = strlen(s1 + 1); 
    N2 = strlen(s2 + 1);
    for(int i = 1; i <= N1; i++) 
        insert(s1[i] - 'a');
    int ans = 0, nowlen = 0, cur = root;
    for(int i = 1; i <= N2; i++, ans = max(ans, nowlen)) {
        int p = s2[i] - 'a';
        if(ch[cur][p]) {nowlen++, cur = ch[cur][p]; continue;}
        for(; cur && !ch[cur][p]; cur = fa[cur]);
        nowlen = len[cur] + 1, cur = ch[cur][p];
    }
    printf("%d\n", ans);
    return 0;
}