UVA-11210-Chinese (Mahjong)

UVA-11210-Chinese (Mahjong)

The 24-page training guide is really drunk. I thought it was an endless loop. Originally, there were too many loop sets. A group of samples, 500 S + recursion, would be reduced as much as possible. nested loops would have killed my complexity. 34*14*13*12*11*10*9*8*7*6*5*4*3*2*1 it's hard to use your mind !!!
Pay special attention to future backtracking. Never forget to call the function and change the variable back.

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        Using namespace std; vector
       
         Have; set
        
          Ans; int ji = 0, kase = 0; bool zuhe (vector
         
           Now) {int flagzong = 0; for (int I = 0; I <now. size (); I ++) if (now [I]! = 99) {flagzong = 1; break;} if (! Flagzong) return true; // finally finished for (int I = 0; I <now. size (); I ++) for (int j = 0; j <now. size (); j ++) for (int k = 0; k <now. size (); k ++) {if (I = j | I = k | j = k | now [I] = 99 | now [j] = 99 | now [k] = 99) continue; int a = now [I], B = now [j], c = now [k]; int flag = 0; if (a = B & a = c) // engraved child {vector
          
            Now1 (now); now1 [I] = 99; now1 [j] = 99; now1 [k] = 99; if (zuhe (now1) return true ;} else if (a> = 0 & a <= 8 & B> = 0 & B <= 8 & c> = 0 & c <= 8) | (a> = 9 & a <= 17 & B> = 9 & B <= 17 & c> = 9 & c <= 17) | (a> = 18 & a <= 26 & B> = 18 & B <= 26 & c> = 18 & c <= 26 )) // shun sub {if (a + 1 = B & B + 1 = c) flag = 1; if (a + 1 = c & c + 1 = B) flag = 1; if (B + 1 = a & a + 1 = c) flag = 1; if (B + 1 = c & c + 1 = a) flag = 1; if (c + 1 = a & a + 1 = B) flag = 1; if (c + 1 = B & B + 1 =) flag = 1; if (flag) {vector
           
             Now1 (now); now1 [I] = 99; now1 [j] = 99; now1 [k] = 99; if (zuhe (now1) return true ;}}} return false;} bool ting (vector
            
              Now) // select two cards for {for (int I = 0; I <now. size (); I ++) for (int j = 0; j <now. size (); j ++) {if (I = j | now [I]! = Now [j]) continue; vector
             
               Temp (now); temp [I] = 99; temp [j] = 99; if (zuhe (temp) return true;} return false;} void solve () {for (int I = 0; I <34; I ++) {int countt = 0; for (int j = 0; j 
                Temp (have); temp. push_back (I); if (ting (temp) ans. insert (I) ;}} void build (string str) {if (str = "DONG") have. push_back (27); if (str = "NAN") have. push_back (28); if (str = "XI") have. push_back (29); if (str = "BEI") have. push_back (30); if (str = "ZHONG") have. push_back (31); if (str = "FA") have. push_back (32); if (str = "BAI") have. push_back (33); if (str [1] = 'T') have. push_back (str [0]-'0'-1); if (str [1] = 's') have. push_back (str [0]-'0' + 8); if (str [1] = 'W') have. push_back (str [0]-'0' + 17);} void print () {set
               
                 : Iterator it; if (ans. empty () {printf ("Not ready \ n"); return ;}for (it = ans. begin (); it! = Ans. end (); it ++) {printf (""); if (* it)> = 0 & * it <= 8) printf ("% dT ", (* it) + 1); if (* it)> = 9 & * it <= 17) printf ("% dS", (* it)-8 ); if (* it)> = 18 & * it <= 26) printf ("% dW", (* it)-17); if (* it) = 27) printf ("DONG"); if (* it) = 28) printf ("NAN"); if (* it) = 29) printf ("XI"); if (* it) = 30) printf ("BEI"); if (* it) = 31) printf ("ZHONG"); if (* it) = 32) printf ("FA"); if (* it) = 33) print F ("BAI");} printf ("\ n"); return;} int main () {string str; while (cin> str & str! = "0") {build (str); ji ++; if (ji % 13 = 0) {solve (); printf ("Case % d :", ++ kase); print (); ji = 0; have. clear (); ans. clear () ;}} return 0 ;}
               
              
             
            
           
          
         
        
       
      
     
    
   
  

The following is the code of Jia Shen:

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Int mj [20], cnt [35]; const char * mahjong [] = {"1 T", "2 T", "3 T", "4 T ", "5 T", "6 T", "7 T", "8 T", "9 T", "1 S", "2 S", "3 S ", "4 S", "5 S", "6 S", "7 S", "8 S", "9 S", "1 W", "2 W ", "3 W", "4 W", "5 W", "6 W", "7 W", "8 W", "9 W", "DONG ", "NAN", "XI", "BEI", "ZHONG", "FA", "BAI"}; int getNum (char * str) {// convert the card to the ID for (int I = 0; I <34; I ++) {if (! Strcmp (mahjong [I], str) return I ;}} int check2 (int n) {for (int I = 0; I <34; I ++) {// engraved subif (cnt [I]> = 3) {if (n = 3) return 1; cnt [I]-= 3; if (check2 (n + 1) return 1; cnt [I] + = 3 ;}}for (int I = 0; I <= 24; I ++) {// subif (I % 9 <= 6 & cnt [I] & cnt [I + 1] & cnt [I + 2]) {if (n = 3) return 1; cnt [I] --; cnt [I + 1] --; cnt [I + 2] --; if (check2 (n + 1) return 1; cnt [I] ++; cnt [I + 1] ++; cn T [I + 2] ++;} return 0;} int check () {for (int I = 0; I <34; I ++) {if (cnt [I]> = 2) {// convert cnt [I]-= 2; if (check2 (0) return 1; cnt [I] + = 2;} return 0;} int main () {char str [3]; int Case = 1; while (scanf ("% s ", str) = 1) {if (strcmp (str, "0") = 0) break; printf ("Case % d:", Case ++ ); mj [0] = getNum (str); for (int I = 1; I <13; I ++) {scanf ("% s", str ); mj [I] = getNum (str); // convert the card to number} int f Lag = 0; for (int I = 0; I <34; I ++) {// enumerated 34 possible conditions: memset (cnt, 0, sizeof (cnt); for (int j = 0; j <13; j ++) {cnt [mj [j] ++; // count the number of occurrences of each card} if (cnt [I]> = 4) continue; // if the number of occurrences of each card appears four times, the card cnt [I] ++ is not considered; if (check () {flag = 1; printf ("% s", mahjong [I]) ;}} if (! Flag) printf ("Not ready/n"); else printf ("/n");} return 0 ;}