CF427D Match & Catch

https://www.luogu.com.cn/problem/CF427D

广义SAM板板题
每个节点分别统计两个串的size，（注意，SAM并不是编号越大的节点深度越大，所以必须要一个dfs来统计）

我是另一种写法，不过无所谓了
code:

#include<bits/stdc++.h>
#define N 200050
using namespace std;
struct A {
    int ch[27], len, fa;
} a[N << 1];
int lst = 1, tot = 1, size[N][2], n;
void insert(int c, int o) {
    int p = lst, np = ++ tot; lst = np;
    a[np].len = a[p].len + 1;
    while(p && !a[p].ch[c]) a[p].ch[c] = np, p = a[p].fa;
    if(!p) a[np].fa = 1;
    else {
        int q = a[p].ch[c];
        if(a[q].len == a[p].len + 1) a[np].fa = q;
        else {
            int clone = ++ tot; a[clone] = a[q];
            a[clone].len = a[p].len + 1; size[clone][0] = size[q][0]; size[clone][1] = size[q][1];
            while(p && a[p].ch[c] == q) a[p].ch[c] = clone, p = a[p].fa;
            a[q].fa = a[np].fa = clone;
        }
    }
   // printf(" %d   *%d %d   ", lst, a[7].fa, size[7][o]);
    int now = lst;
    while(now && size[now][o] <= 1) size[now][o] ++, now = a[now].fa;//, printf("(%d %d) ", now, size[now][o]); printf("\n");
  //  printf(" %d   *%d %d   \n", now, a[7].fa, size[7][o]);

}
char st[N], stt[N];
int main() {
    scanf("%s", st + 1);
    int n = strlen(st + 1);
    for(int i = 1; i <= n; i ++) insert(st[i] - 'a', 0);
  //  for(int i = 1; i <= tot; i ++) printf("%d ", size[i][0]); printf("\n");
    scanf(" %s", stt + 1);
    n = strlen(stt + 1);
    lst = 1;
    for(int i = 1; i <= n; i ++) insert(stt[i] - 'a', 1);
//    for(int i = 1; i <= tot; i ++) printf("%d ", size[i][0]); printf("\n");
  //  for(int i = 1; i <= tot; i ++) printf("%d ", size[i][1]); printf("\n");

    /* for(int i = tot; i >= 1; i --) {
        size[a[i].fa][0] += size[i][0];
        size[a[i].fa][1] += size[i][1];
    }*/
    int ans = n + 1;
    for(int i = 2; i <= tot; i ++) {
        if(size[i][0] == 1 && size[i][1] == 1) {
            ans = min(ans, a[a[i].fa].len + 1);
         //   printf("%d %d   %d\n", i, ans, n);
        }
    }
    if(ans == n + 1) ans = -1;
    printf("%d", ans);
    return 0;
}