1204 word puzzles

When you are a beginner, you can use the code of someone else to modify the template.

When I used query () to query a suitable solution, I returned the result. After wa n times, I made such a low-level error, so I should not.

I have benefited a lot from reading other people's code this time, so I can record it

# Include <iostream> <br/> # include <stdio. h> <br/> # include <string. h> <br/> using namespace STD; <br/> # define txtl 1005 </P> <p> // when a subtree node is inserted, <br/> // The queue used in the search for the mismatched pointer is simulated using an array. <br/> const int kind = 26; // Number of subtree <br/> struct node <br/> {<br/> node * fail; <br/> node * Next [kind]; <br/> int count; // record the number of complete words with the current prefix <br/> node () <br/>{< br/> fail = NULL; <br/> COUNT =-1; <br/> memset (next, 0, sizeof (next); <br/>}< br/>} * Q [1000001]; // The queue used to find the mismatched pointer <br/> struct cword {<br/> char data [1001]; <br/> int wordlen; <br/> int row, Col; <br/> char Ori; <br/>}; <br/> const int dir [] [2] = {-1, 0}, {-1, 1}, {0, 1}, {1, 1}, {1, 0}, {1,-1}, {0, -1 },{-1,-1 }}; <br/> char WP [txtl] [txtl]; <br/> cword words [txtl]; <br/> int row, Col, N; <br/> void insert (node * root, char * STR, int ID) <br/> {<br/> node * P = root; <br/> int I = 0, index; <br/> while (STR [I]) <br/>{< br/> Index = STR [I]-'A'; <br/> If (p-> next [Index] = NULL) <br/> P-> next [Index] = new node (); <br/> P = p-> next [Index]; <br/> I ++; <br/>}< br/> P-> COUNT = ID; <br/>}< br/> // search for failure pointers <br/> void build_ac_automation (node * root) <br/>{< br/> int I; <br/> int head, tail; <br/> root-> fail = NULL; <br/> head = 0; <br/> tail = 0; <br/> q [tail ++] = root; <br/> while (Head! = Tail) <br/>{< br/> node * temp = Q [head ++]; // gets the first element of the queue <br/> node * P = NULL; <br/> for (I = 0; I <kind; I ++) <br/>{< br/> If (temp-> next [I]! = NULL) // find the failure pointer of the current subtree <br/>{< br/> P = temp-> fail; <br/> while (P! = NULL) <br/>{< br/> If (p-> next [I]! = NULL) // find the failure pointer <br/>{< br/> temp-> next [I]-> fail = p-> next [I]; <br/> break; <br/>}< br/> P = p-> fail; <br/>}< br/> If (P = NULL) // cannot be obtained. The current subtree failure pointer is the root <br/> temp-> next [I]-> fail = root; <br/> q [tail ++] = temp-> next [I]; // enter the current subtree <br/>}< br/> // query that STR contains N keywords number of <br/> void query (node * root, char * STR, Int & wordid, int IR, int IC, int di) <br/>{< br/> int I = 0, CNT = 0, index, Len; <br/> Len = strlen (St R); <br/> node * P = root; <br/> while (STR [I]) <br/>{< br/> Index = STR [I]-'A'; <br/> while (p-> next [Index] = NULL & P! = Root) // mismatch <br/> P = p-> fail; <br/> P = p-> next [Index]; <br/> If (P = NULL) // The mismatched pointer is the root <br/> P = root; <br/> node * temp = P; <br/> while (temp! = Root) // mark the accessed point in this process <br/>{< br/> If (temp-> count! =-1) {<br/> wordid = temp-> count; <br/> // return I; a suitable solution is returned after a query is found, wa has made such a low-level mistake n times, so it is not necessary <br/> words [wordid]. row = IR + (I-words [wordid]. wordlen + 1) * dir [di] [0]; <br/> words [wordid]. col = IC + (I-words [wordid]. wordlen + 1) * dir [di] [1]; <br/> words [wordid]. ori = di + 'a'; <br/>}< br/> temp = temp-> fail; <br/>}< br/> I ++; <br/>}< br/> inline int inbound (int r, int C) {<br/> return 0 <= R & R <row & 0 <= C & C <Col; <br/>}< br/> void findsub (node * root, int IR, int IC, int di) {<br/> int JS, wordid, RS; <br/> char ss [txtl]; <br/> for (JS = 0; inbound (IR + js * dir [di] [0], IC + js * dir [di] [1]); ++ JS) {<br/> SS [JS] = WP [IR + js * dir [di] [0] [IC + js * dir [di] [1]; <br/>}< br/> SS [JS] = '/0'; <br/> query (root, SS, wordid, IR, IC, DI ); <br/>}< br/> int main () <br/> {<br/> scanf ("% d", & Row, & Col, & N); <br/> node * root = new node (); <br/> for (INT I = 0; I <row; ++ I) {<br/> scanf ("% s", WP [I]); <br/>}< br/> for (INT I = 0; I <N; ++ I) <br/>{< br/> scanf ("% s", words [I]. data); <br/> words [I]. wordlen = strlen (words [I]. data); <br/> insert (root, words [I]. data, I); <br/>}< br/> // find the failure pointer <br/> build_ac_automation (Root); <br/> for (INT I = 0; I <n; ++ I) {<br/> words [I]. col = words [I]. row = words [I]. ori =-100; <br/>}< br/> for (Int J = col-1; j> = 0; -- j) {<br/> findsub (root, 0, J, 3); <br/> findsub (root, 0, J, 4); <br/> findsub (root, 0, J, 5 ); <br/> findsub (root, row-1, J, 7); <br/> findsub (root, row-1, J, 0 ); <br/> findsub (root, row-1, J, 1); <br/>}< br/> for (INT I = row-1; I >= 0; -- I) {<br/> findsub (root, I, 0, 1); <br/> findsub (root, I, 0, 2 ); <br/> findsub (root, I, 0, 3); <br/> findsub (root, I, col-1, 5); <br/> findsub (root, i, col-1, 6); <br/> findsub (root, I, col-1, 7 ); <br/>}< br/> for (INT I = 0; I <n; ++ I) {<br/> printf ("% d % C/N", words [I]. row, words [I]. col, words [I]. ori); </P> <p >}< br/> return 0; <br/>}< br/>