Secret Passwords

int n,fa[50];
char s[1001];
int vis[200001], ans;
int fin(int x){
    return x == fa[x] ? x : fa[x] = fin(fa[x]);
}
int main() {
    cin >> n;
    for (int i = 1; i <= 30; i++) {
        fa[i] = i;  //各自为一个集合
    }
    for (int i = 1; i <= n; i++) {
        cin >> s;
        int l = strlen(s);
        for (int j = 0; j < l - 1; j++) {
            vis[s[j] - ‘a‘ + 1] = 1;
            int x = fin(s[j] - ‘a‘ + 1), y = fin(s[j + 1] - ‘a‘ + 1);
            if (x != y) {
                fa[x] = y;  //若在一个单词内，祖宗结点不一致，则合并祖宗结点
            }
            vis[s[l - 1] - ‘a‘ + 1] = 1;
        }
    }
    for (int i = 1; i <= 26; i++){
        if (vis[i] && fa[i] == i) {  //最后在n个单词中，只要单词与单词之间有字母相同，则祖宗结点一致
            ans++;  //统计祖宗结点未变的个数并且已访问的
        }
    }
    cout << ans << endl;
    return 0;
}