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131 | #include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int N = 2005;
char s[N], t[N];
int na[N][26], nb[N][26], nxt[26];
int n, m, a[N], b[N], tot = 1, p = 1, f[N][N << 1];
struct SAM {
int par, ch[26], len;
} sam[N << 1];
void insert(int x) {
int np = ++tot; // 新节点
sam[np].len = sam[p].len + 1;
while (p && !sam[p].ch[x]) {
sam[p].ch[x] = np;
p = sam[p].par;
}
if (p == 0)
sam[np].par = 1;
else {
int q = sam[p].ch[x];
if (sam[q].len == sam[p].len + 1)
sam[np].par = q;
else {
int nq = ++tot;
sam[nq].len = sam[p].len + 1;
memcpy(sam[nq].ch, sam[q].ch, sizeof(sam[q].ch));
sam[nq].par = sam[q].par;
sam[q].par = sam[np].par = nq;
while (p && sam[p].ch[x] == q) {
sam[p].ch[x] = nq;
p = sam[p].par;
}
}
}
p = np;
}
int main() {
scanf("%s%s", s + 1, t + 1);
n = strlen(s + 1);
m = strlen(t + 1);
for (int i = 1; i <= n; ++i) a[i] = s[i] - 'a';
for (int i = 1; i <= m; ++i) b[i] = t[i] - 'a';
for (int i = 1; i <= m; ++i) insert(b[i]);
// nxt[S[i]]<-i
for (int i = 0; i < 26; ++i) nxt[i] = n + 1;
for (int i = n; i >= 0; --i) {
memcpy(na[i], nxt, sizeof(nxt));
nxt[a[i]] = i;
}
for (int i = 0; i < 26; ++i) nxt[i] = m + 1;
for (int i = m; i >= 0; --i) {
memcpy(nb[i], nxt, sizeof(nxt));
nxt[b[i]] = i;
}
// 四种情况计算答案
// 1
int ans = N;
for (int l = 1; l <= n; ++l) {
for (int r = l, u = 1; r <= n; ++r) {
u = sam[u].ch[a[r]];
if (!u) {
ans = min(ans, r - l + 1);
break;
}
}
}
printf("%d\n", ans == N ? -1 : ans);
// 2
ans = N;
for (int l = 1; l <= n; ++l) {
for (int r = l, u = 0; r <= n; ++r) {
u = nb[u][a[r]];
if (u == m + 1) {
ans = min(ans, r - l + 1);
break;
}
}
}
printf("%d\n", ans == N ? -1 : ans);
// 3
for (int i = n; i >= 0; --i) {
for (int j = 1; j <= tot; ++j) {
f[i][j] = N;
for (int c = 0; c < 26; ++c) {
int u = na[i][c];
int v = sam[j].ch[c];
if (u <= n) f[i][j] = min(f[i][j], f[u][v] + 1);
}
}
}
printf("%d\n", f[0][1] == N ? -1 : f[0][1]);
// 4
memset(f, 0, sizeof(f));
for (int i = n; i >= 0; --i) {
for (int j = 0; j <= m; ++j) {
f[i][j] = N;
for (int c = 0; c < 26; ++c) {
int u = na[i][c];
int v = nb[j][c];
if (u <= n) f[i][j] = min(f[i][j], f[u][v] + 1);
}
}
}
printf("%d\n", f[0][0] == N ? -1 : f[0][0]);
return 0;
}
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