Poj 1077 Digital 8

Poj 1077 Digital 8

This topic took me one night to read a solution. The final result was a tragedy. I only passed it on poj. On hdu, TLE is because hdu contains multiple groups of data.

Data, and poj is a group of data... Unfortunately, the learning method is a little inefficient...

However, the print path method can be used for reference to recursively print the path from the end point to the start point...

Code:

 

# Include
 
  
# Include
  
   
# Include
   
    
# Include
    
     
Using namespace std; # define N 371_int first [9]; int last [9] = {1, 2, 4, 5, 6, 7, 8, 0}; int f [9]; int visit [N]; int p [N] [9]; // an array of saved status int dist [N]; // record path const int dx [] = {-,}; const int dy [] = {0,-, 0 }; char dir [4] = {'D', 'R', 'l', 'U'}; void init () // It seems to be a magic horse Dongdong {f [0] = 1; for (int I = 1; I <10; I ++) f [I] = f [I-1] * I;} int hash1 (int * a) // This is the number of digits 0-9 in the full arrangement, used to determine the {int I, j; int s = 0; for (I = 0; I <9; I ++) {int cnt = 0; for (j = I + 1; j <9; j ++) {if (a [j] <a [I]) cnt ++ ;} s + = f [8-i] * cnt;} return s;} int bfs (int fir, int las) {int I, cur; memset (visit, 0, sizeof (visit); memset (dist, 0, sizeof (dist); memset (p, 0, sizeof (p); visit [fir] = 1; dist [fir] = 0; for (I = 0; I <9; I ++) p [fir] [I] = first [I]; queue
     
      
Que; que. push (fir); while (! Que. empty () {int temp = que. front (); que. pop (); if (temp = las) return dist [las]; for (cur = 0; cur <9; cur ++) {if (p [temp] [cur] = 0) break;} int x = cur/3; int y = cur % 3; int change [9]; for (I = 0; I <4; I ++) {int xx = x + dx [I]; int yy = y + dy [I]; if (xx> = 0) & (yy> = 0) & (xx <3) & (yy <3) {memcpy (change, p [temp], sizeof (change); change [cur] = p [temp] [xx * 3 + yy]; change [xx * 3 + yy] = 0; int change1 = hash1 (change); if (visit [change1] = 0) {visit [change1] = 1; dist [change1] = dist [temp] + 1; que. push (change1); memcpy (p [change1], change, sizeof (change) ;}}} return-1 ;}void print_path (int fir, int las) // recursively print the path from the end point {if (las = fir) return; int cur; int change [9]; for (cur = 0; cur <9; cur ++) {if (p [las] [cur] = 0) break;} int x = cur/3; int y = cur % 3; for (int I = 0; I <4; I ++) {int xx = x + dx [I]; int yy = y + dy [I]; if (xx> = 0) & (yy> = 0) & (xx <3) & (yy <3) {memcpy (change, p [las], sizeof (change )); change [cur] = p [las] [xx * 3 + yy]; change [xx * 3 + yy] = 0; int change1 = hash1 (change ); if (dist [las] = dist [change1] + 1) & visit [change1]) {print_path (fir, change1); printf (% c, dir [I]); break ;}}return ;}int main () {init (); int I, k = 0; char a [30]; while (gets (a) {// getchar (); k = 0; int len = strlen (a); for (I = 0; I
      
        = '1') & (a [I] <= '8') first [k ++] = a [I]-'0 '; else if (a [I] = 'X') first [k ++] = 0 ;}// for (I = 0; I <9; I ++) // printf (% d, first [I]); puts (); int fir = hash1 (first ); // find the fully arranged bits corresponding to a sequence. int las = hash1 (last); int ans = bfs (fir, las); // printf (% d, ans); if (ans <0) printf (unsolvable); elseprint_path (fir, las); puts ();} return 0 ;}