poj 1998 Lloyd Fifteen Puzzle

poj 1998 Lloyd Fifteen Puzzle

#include<iostream>#include<string>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;#define VEC(l) (l.b-l.a)#define DBG 1#define sz(c) ((int)(c).size())#define    forl(i, a, b) for(int i = (a); i <  (b); ++i)#define   forle(i, a, b) for(int i = (a); i <= (b); ++i)#define    forg(i, a, b) for(int i = (a); i >  (b); --i)#define   forge(i, a, b) for(int i = (a); i >= (b); --i)#define   forlc(i, a, b) for(int i##_b = (b), i = (a); i <  i##_b; ++i)#define  forlec(i, a, b) for(int i##_b = (b), i = (a); i <= i##_b; ++i)#define   forgc(i, a, b) for(int i##_b = (b), i = (a); i >  i##_b; --i)#define  forgec(i, a, b) for(int i##_b = (b), i = (a); i >= i##_b; --i)#define  forall(i, v   )  forl(i, 0, sz(v))#define forallc(i, v   ) forlc(i, 0, sz(v))#define  forlla(i, v   ) forge(i, sz(v)-1, 0)#define   forls(i, n, a, b) for(int i = a; i != b; i = n[i])#define rep(n)  for(int    repp = 0; repp <    (n); ++repp)#define repc(n) for(int repp_b = (n), repp = 0; repp < repp_b; ++repp)#define rst(a, v) memset(a, v, sizeof a)#define cpy(a, b) memcpy(a, b, sizeof a)#define rstn(a, v, n) memset(a, v, (n)*sizeof((a)[0]))#define cpyn(a, b, n) memcpy(a, b, (n)*sizeof((a)[0]))#define ast(b) if(DBG && !(b)) { printf("%d!!|n", __LINE__); while(1) getchar(); }#define dout DBG && cout << __LINE__ << ">>| "#define pr(x) #x"=" << (x) << " | "#define mk(x) DBG && cout << __LINE__ << "**| "#x << endl#define pra(arr, a, b) if(DBG) {     dout<<#arr"[] | ";     forlec(i, a, b) cout<<"["<<i<<"]="<<arr[i]<<" |"<<((i-(a)+1)%13?" ":"n");     if(((b)-(a)+1)%13) puts("");   }#define rd(type, x) type x; cin >> xinline int     rdi() { int d; scanf("%d", &d); return d; }inline char    rdc() { scanf(" "); return getchar(); }inline string  rds() { rd(string, s); return s; }inline double rddb() { double d; scanf("%lf", &d); return d; }template<class T> inline bool updateMin(T& a, T b) { return a>b? a=b, true: false; }template<class T> inline bool updateMax(T& a, T b) { return a<b? a=b, true: false; }const int N = 1024;const int di[4] = {-1,1,0,0};const int dj[4] = {0,0,-1,1};const char* ds[4] = {"dolu", "nahoru", "doprava", "doleva"};int r, c;int oi, oj;int a[N][N];bool in_bounds(int i, int j) {  return 0<=i&&i<r && 0<=j&&j<c;}void mov(int num) {  int dir;  for(dir = 0; dir < 4; ++dir) {    int ni = oi+di[dir];    int nj = oj+dj[dir];    if(in_bounds(ni, nj) && a[ni][nj]==num) {      printf("Kamen %d presunut %s.n", num, ds[dir]);      swap(a[oi][oj], a[ni][nj]);      oi = ni;      oj = nj;      break;    }  }  if(dir == 4) printf("Neplatny tah kamenem %d.n", num);}void print_a() {  forl(i, 0, r)    forl(j, 0, c)      printf("%d%c", a[i][j], j<c-1?' ':'n');}int main() {  repc(rdi()) {    scanf("%d %d", &r, &c);    forl(i, 0, r)      forl(j, 0, c) {        scanf("%d", a[i]+j);        if(a[i][j] == 0) oi = i, oj = j;      }    static int cas;    printf("Skladacka cislo %d:n", ++cas);    for(int num; scanf("%d", &num)==1 && num; mov(num)) {}    print_a();    puts("");  }  return 0;}