IME Starters Try-outs 2018 J. JHADCBEIGF

Posted 2020-12-03 bigtom

J. JHADCBEIGF

time limit per test

1.0 s

memory limit per test

256 MB

input

standard input

output

standard output

The members from Incomprehensive Message Encryptors (IME) are learning a new cipher. This technique, although kinda weird, consists of using a key $\mathrm{kk that\; is\; a\; permutation\; of\; the\; alphabet.\; In\; other\; words,\; if\; the\; alphabet \Sigma \Sigma is\; a\; sequence\; of nn symbols a1,a2,\dots ,ana1,a2,\dots ,an,\; the\; key kk is\; a\; sequence\; of nn symbols ap1,ap2,...,apnap1,ap2,...,apn,\; for\; some\; permutation p1,p2,\dots ,pnp1,p2,\dots ,pn of 1,2,...,n1,2,...,n.}$

They learned to use the key to encrypt a string $\mathrm{SS by\; the\; following\; technique:\; for\; each ii in 1,2,\dots ,n1,2,\dots ,n,\; they\; switch\; each\; occurrence\; of\; the\; character aiai in\; the\; string SS by\; the\; corresponding\; character apiapi in\; the\; key kk.\; This\; substitution\; occurs\; for\; all\; characters\; at\; once,\; so\; every\; substitution\; doesn’t\; influence\; the\; other\; ones.}$

For example, if $\mathrm{\Sigma =\{A,B,C,D,E\}\Sigma =\{A,B,C,D,E\} and k=\{E,D,C,A,B\}k=\{E,D,C,A,B\},\; the\; encryption\; will\; change\; every\; symbol AA to EE, BB to DD, CC to CC, DD to AA and EE to BB.\; So\; the\; string ABCBADEABCBADE after\; encryption\; turns\; into EDCDEABEDCDEAB.}$

After learning the encryption, the members received $\mathrm{mm tasks\; to\; execute.\; There\; are\; three\; types\; of\; tasks:}$

• $\mathrm{11 ss:\; add\; the\; string ss to\; the\; dictionary;}$
• $\mathrm{22:\; for\; each\; string\; currently\; in\; the\; dictionary,\; change\; it\; by\; its\; encrypted\; version;}$
• $\mathrm{33 ss:\; check\; if\; the\; string ss is\; prefix\; of\; some\; string\; in\; the\; dictionary.}$

Because of how complex it is to encrypt it manually, and they are not really good at programming (by their incomprehensiveness), they need someone to solve these tasks for them. That’s why they gave you this responsibility.

The alphabet is always the first $\mathrm{nn uppercase\; English\; letters\; in\; alphabetical\; order.}$

Input

The first line of input contains two integers $\mathrm{nn, mm ( 1\le n\le 261\le n\le 26, 0\le m\le 1050\le m\le 105)\; —\; the\; size\; of\; the\; alphabet\; and\; the\; number\; of\; tasks.}$

The second line contains one string $\mathrm{kk (|k|=n|k|=n)\; —\; the\; initial\; key.\; It‘s\; guaranteed\; that kk is\; a\; permutation\; of\; the\; alphabet.}$

The next $\mathrm{mm lines\; contains,\; each,\; one\; integer titi (1\le ti\le 31\le ti\le 3)\; —\; the\; type\; of\; the\; task ii.}$

If ${\mathrm{titi is 11,\; the\; line\; also\; contains\; one\; string sisi (1\le |si|\le 5\times 1051\le |si|\le 5\times 105)\; —\; the\; string\; to\; be\; added\; to\; the\; dictionary.}}^{}$

If ${\mathrm{titi is 33,\; the\; line\; also\; contains\; one\; string sisi (1\le |si|\le 5\times 1051\le |si|\le 5\times 105)\; —\; the\; string\; to\; be\; checked\; against\; the\; dictionary.\; It‘s\; guaranteed\; that sisionly\; contains\; symbols\; from\; the\; alphabet.}}^{}$

It‘s guaranteed that there‘s at least one task of type $\mathrm{33 and\; that \sum |si|\le 5 times105\sum |si|\le 5 times105.}$

Output

For each task of type $\mathrm{33,\; print\; “YES”\; if\; the\; string\; is\; a\; prefix\; of\; some\; string\; in\; the\; dictionary,\; and\; “NO”\; otherwise.}$

Examples

input

Copy

5 10
ECABD
1 ABACABA
1 EAAE
2
1 AADE
3 ECEAECE
3 DED
2
3 EAAE
3 BDDB
3 AB

output

Copy

YES
NO
NO
YES
NO

input

Copy

5 12
BCDEA
1 AAAAA
3 AAAAA
2
3 AAAAA
2
3 AAAAA
2
3 AAAAA
2
3 AAAAA
2
3 AAAAA

output

Copy

YES
NO
NO
NO
NO
YES

思路：维护一下整颗字典树的变换次数，插入时加入这个次数变换前的字符串，查询也调用k次变换前的串。于是就是个字典树裸题了。
//指针式，不要乱delete，会把指针指向的那块区域全都删掉

  1 #include <iostream>
  2 #include <fstream>
  3 #include <sstream>
  4 #include <cstdlib>
  5 #include <cstdio>
  6 #include <cmath>
  7 #include <string>
  8 #include <cstring>
  9 #include <algorithm>
 10 #include <queue>
 11 #include <stack>
 12 #include <vector>
 13 #include <set>
 14 #include <map>
 15 #include <list>
 16 #include <iomanip>
 17 #include <cctype>
 18 #include <cassert>
 19 #include <bitset>
 20 #include <ctime>
 21 
 22 using namespace std;
 23 
 24 #define pau system("pause")
 25 #define ll long long
 26 #define pii pair<int, int>
 27 #define pb push_back
 28 #define mp make_pair
 29 #define clr(a, x) memset(a, x, sizeof(a))
 30 
 31 const double pi = acos(-1.0);
 32 const int INF = 0x3f3f3f3f;
 33 const int MOD = 1e9 + 7;
 34 const double EPS = 1e-9;
 35 
 36 /*
 37 #include <ext/pb_ds/assoc_container.hpp>
 38 #include <ext/pb_ds/tree_policy.hpp>
 39 
 40 using namespace __gnu_pbds;
 41 tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
 42 */
 43 
 44 int n, m;
 45 char k0[37], k[37], s[500015];
 46 char con[2][27];
 47 void pre() {
 48     for (int i = 1; i <= n; ++i) {
 49         k[k0[i] - ‘A‘ + 1] = ‘A‘ + i - 1;
 50         con[0][i] = ‘A‘ + i - 1;
 51     }
 52 }
 53 struct Trie {
 54     Trie *nex[27];
 55     void ini() {
 56         for (int i = 1; i <= 26; ++i) {
 57             nex[i] = NULL;
 58         }
 59     }
 60 };
 61 void Insert(Trie *T, char s[]) {
 62     int l = strlen(s + 1);
 63     for (int i = 1; i <= l; ++i) {
 64         int x = s[i] - ‘A‘ + 1;
 65         if (NULL == T -> nex[x]) {
 66             T -> nex[x] = new Trie;
 67             T = T -> nex[x];
 68             T -> ini();
 69         } else {
 70             T = T -> nex[x];
 71         }
 72     }
 73 }
 74 bool query(Trie *T, char s[]) {
 75     int l = strlen(s + 1);
 76     for (int i = 1; i <= l; ++i) {
 77         int x = s[i] - ‘A‘ + 1;
 78         if (NULL == T -> nex[x]) {
 79             return false;
 80         } else {
 81             T = T -> nex[x];
 82         }
 83     }
 84     return true;
 85 }
 86 int main() {
 87     scanf("%d%d%s", &n, &m, k0 + 1);
 88     pre();
 89     Trie *T = new Trie;
 90     T -> ini();
 91     int cnt = 0;
 92     while (m--) {
 93         int t;
 94         scanf("%d", &t);
 95         if (1 == t) {
 96             scanf("%s", s + 1);
 97             int l = strlen(s + 1);
 98             for (int i = 1; i <= l; ++i) {
 99                 s[i] = con[cnt][s[i] - ‘A‘ + 1];
100             }
101             Insert(T, s);
102         } else if (2 == t) {
103             cnt ^= 1;
104             for (int i = 1; i <= n; ++i) {
105                 con[cnt][i] = k[con[cnt ^ 1][i] - ‘A‘ + 1];
106             }
107         } else {
108             scanf("%s", s + 1);
109             int l = strlen(s + 1);
110             for (int i = 1; i <= l; ++i) {
111                 s[i] = con[cnt][s[i] - ‘A‘ + 1];
112             }
113             puts(query(T, s) ? "YES" : "NO");
114         }
115     }
116     return 0;
117 }

View Code

　// 数组式

  1 #include <iostream>
  2 #include <fstream>
  3 #include <sstream>
  4 #include <cstdlib>
  5 #include <cstdio>
  6 #include <cmath>
  7 #include <string>
  8 #include <cstring>
  9 #include <algorithm>
 10 #include <queue>
 11 #include <stack>
 12 #include <vector>
 13 #include <set>
 14 #include <map>
 15 #include <list>
 16 #include <iomanip>
 17 #include <cctype>
 18 #include <cassert>
 19 #include <bitset>
 20 #include <ctime>
 21 
 22 using namespace std;
 23 
 24 #define pau system("pause")
 25 #define ll long long
 26 #define pii pair<int, int>
 27 #define pb push_back
 28 #define mp make_pair
 29 #define clr(a, x) memset(a, x, sizeof(a))
 30 
 31 const double pi = acos(-1.0);
 32 const int INF = 0x3f3f3f3f;
 33 const int MOD = 1e9 + 7;
 34 const double EPS = 1e-9;
 35 
 36 /*
 37 #include <ext/pb_ds/assoc_container.hpp>
 38 #include <ext/pb_ds/tree_policy.hpp>
 39 
 40 using namespace __gnu_pbds;
 41 tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
 42 */
 43 
 44 int n, m;
 45 char k0[37], k[37], s[2000015];
 46 char con[2][27];
 47 void pre() {
 48     for (int i = 1; i <= n; ++i) {
 49         k[k0[i] - ‘A‘ + 1] = ‘A‘ + i - 1;
 50         con[0][i] = ‘A‘ + i - 1;
 51     }
 52 }
 53 int nex[2000015][27], cnt_node;
 54 void Insert(int sta, char s[]) {
 55     int l = strlen(s + 1);
 56     for (int i = 1; i <= l; ++i) {
 57         int x = s[i] - ‘A‘ + 1;
 58         if (!nex[sta][x]) {
 59             nex[sta][x] = ++cnt_node;
 60             sta = nex[sta][x];
 61         } else {
 62             sta = nex[sta][x];
 63         }
 64     }
 65 }
 66 bool query(int sta, char s[]) {
 67     int l = strlen(s + 1);
 68     for (int i = 1; i <= l; ++i) {
 69         int x = s[i] - ‘A‘ + 1;
 70         if (!nex[sta][x]) {
 71             return false;
 72         } else {
 73             sta = nex[sta][x];
 74         }
 75     }
 76     return true;
 77 }
 78 int main() {
 79     scanf("%d%d%s", &n, &m, k0 + 1);
 80     pre();
 81     int cnt = 0;
 82     while (m--) {
 83         int t;
 84         scanf("%d", &t);
 85         if (1 == t) {
 86             scanf("%s", s + 1);
 87             int l = strlen(s + 1);
 88             for (int i = 1; i <= l; ++i) {
 89                 s[i] = con[cnt][s[i] - ‘A‘ + 1];
 90             }
 91             Insert(0, s);
 92         } else if (2 == t) {
 93             cnt ^= 1;
 94             for (int i = 1; i <= n; ++i) {
 95                 con[cnt][i] = k[con[cnt ^ 1][i] - ‘A‘ + 1];
 96             }
 97         } else {
 98             scanf("%s", s + 1);
 99             int l = strlen(s + 1);
100             for (int i = 1; i <= l; ++i) {
101                 s[i] = con[cnt][s[i] - ‘A‘ + 1];
102             }
103             puts(query(0, s) ? "YES" : "NO");
104         }
105     }
106     return 0;
107 }

View Code