【SP1812】LCS2 - Longest Common Substring II

【SP1812】LCS2 - Longest Common Substring II

题面

洛谷

题解

你首先得会做这题。

然后就其实就很简单了，

你在每一个状态\(i\)打一个标记\(f[i]\)表示状态\(i\)能匹配到最长的子串长度，

显然\(f[i]\)可以上传给\(f[i.fa]\)。

然后去每个串和第\(1\)个串\(f\)的最小值的最大值即可。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAX_N = 1e5 + 5;
struct Node { int ch[26], fa, len; } t[MAX_N << 1];
int tot = 1, lst = 1;
void extend(int c) {
	++tot, t[lst].ch[c] = tot;
	t[tot].len = t[lst].len + 1;
	int p = t[lst].fa; lst = tot;
	while (p && !t[p].ch[c]) t[p].ch[c] = tot, p = t[p].fa;
	if (!p) return (void)(t[tot].fa = 1);
	int q = t[p].ch[c];
	if (t[q].len == t[p].len + 1) return (void)(t[tot].fa = q);
	int _q = ++tot; t[_q] = t[q];
	t[_q].len = t[p].len + 1, t[q].fa = t[tot - 1].fa = _q;
	while (p && t[p].ch[c] == q) t[p].ch[c] = _q, p = t[p].fa;
}
char a[MAX_N];
int N, A[MAX_N << 1], f[MAX_N << 1], g[MAX_N << 1], bln[MAX_N << 1];
int main () {
#ifndef ONLINE_JUDGE
    freopen("cpp.in", "r", stdin);
#endif
	scanf("%s", a + 1);
	N = strlen(a + 1);
	for (int i = 1; i <= N; i++) extend(a[i] - 'a');
	for (int i = 1; i <= tot; i++) ++bln[t[i].len];
	for (int i = 1; i <= tot; i++) bln[i] += bln[i - 1];
	for (int i = 1; i <= tot; i++) A[bln[t[i].len]--] = i;
	for (int i = 1; i <= tot; i++) g[i] = t[i].len;
	while (scanf("%s", a + 1) != EOF) {
		N = strlen(a + 1);
		for (int i = 1; i <= tot; i++) f[i] = 0;
		for (int v = 1, l = 0, i = 1; i <= N; i++) {
			while (v && !t[v].ch[a[i] - 'a']) v = t[v].fa, l = t[v].len;
			if (!v) v = 1, l = 0;
			if (t[v].ch[a[i] - 'a']) ++l, v = t[v].ch[a[i] - 'a'];
			f[v] = max(f[v], l);
		}
		for (int i = tot; i; i--) f[t[i].fa] = max(f[t[i].fa], f[i]);
		for (int i = 1; i <= tot; i++) g[i] = min(g[i], f[i]);
	}
	printf("%d\n", *max_element(&g[1], &g[tot + 1]));
    return 0;
}