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
", ans);
return 0;
}