Nightmare HDU 1072 BFS (the BFS that can go back, hash is a bit handsome)

/* <Br/> it seems that it is really a lot of water. If you have written a question in the past, TLE <br/> only remember to get a 6 in hash, which will be AC, but it does not mean that you will <br/> This hash [SX] [sy] = 6 is really a good method <br/> */# include <iostream> <br/> # include <cstdio> <br/> # include <cstring> <br/> # include <queue> <br/> using namespace STD; <br/> const int n = 12; <br/> int hash [N] [N]; <br/> int map [N] [N]; <br/> struct node <br/> {<br/> int X, Y, step; <br/> node (){}; <br/> node (int xx, int YY, int st): X (XX), y (yy), St EP (ST) {}; <br/>}; <br/> int DX [] = {0, 0, 1,-1 }; <br/> int dy [] = {1,-1, 0, 0 };< br/> int n, m; <br/> void BFS (INT Sx, int Sy, int ex, int ey) <br/>{< br/> node now, next; <br/> queue <node> q; <br/> q. push (node (sx, Sy, 0); <br/> hash [SX] [sy] = 6; // the start time is 6 <br/> while (! Q. empty () <br/>{< br/> now = Q. front (); <br/> q. pop (); <br/> for (INT I = 0; I <4; I ++) <br/> {<br/> next = now; <br/> next. X = now. X + dx [I]; <br/> next. y = now. Y + dy [I]; <br/> If (next. x <0 | next. x> = n | next. Y <0 | next. y> = M | hash [now. x] [now. y] <= 1 | map [next. x] [next. y] = 0) <br/> continue; <br/> next. step = now. step + 1; <br/> If (Map [next. x] [next. y] = 3) <br/>{< br/> printf ("% d/N", next. step); <br/> return; <br/>}< br/> If (hash [next. x] [next. y] <pash [now. x] [now. y]-1) <br/> {// The next point enters the queue only when the explosion time is shorter than the explosion time of now. <br/> hash [next. x] [next. y] = hash [now. x] [now. y]-1; <br/> If (Map [next. x] [next. y] = 4) <br/>{< br/> hash [next. x] [next. y] = 6; <br/> // next. time = 6; <br/>}< br/> q. push (next); <br/>}</P> <p >}< br/> printf ("-1/N "); <br/> return; <br/>}< br/> int main () <br/>{< br/> // freopen ("hxsh. in "," r ", stdin); // do not copy such a file all day. <br/> int T; <br/> scanf ("% d", & T); <br/> while (t --) <br/>{< br/> scanf ("% d", & N, & M); <br/> int Sx, Sy, ex, ey; <br/> for (INT I = 0; I <n; I ++) <br/> for (Int J = 0; j <m; j ++) <br/> {<br/> hash [I] [J] = 0; <br/> scanf ("% d", & map [I] [J]); <br/> If (Map [I] [J] = 2) <br/> {<br/> SX = I; <br/> Sy = J; <br/>}< br/> If (Map [I] [J] = 3) <br/>{< br/> ex = I; <br/> ey = J; <br/>}< br/> BFS (sx, Sy, ex, ey ); <br/>}< br/>} 

/*
The method draws on the hdoj Forum and has not dealt with any questions that can be taken back. Therefore, it is very tricky to deal,
The memory burst for the first time. For the output is-1, I initially considered a problem, but I don't know how to solve it.
Later, I came to the Forum and thought it was because of the endless loop caused by my code.
The deformation of BFS really makes people feel a little bit ..
*/
# Include <iostream> // 2399633 2010-04-29 19:31:23 accepted 1072 0 Ms 292 K 1915 B C ++ regret
# Include <cstdio>
# Include <cmath>
# Include <queue>
Using namespace STD;

Int Si, SJ;
Int EI, EJ;
Int num;
Int n, m;
Char map [10] [10];
Int hash [10] [10]; // This hash is a bit handsome...
Int dir [4] [2] = {1, 0}, {-1, 0}, {0, 1}, {0,-1 }};
Struct Node
{
Int time;
Int X;
Int y;
// Int count;
};

Int BFS (INT Sx, int SY)
{
Node N, P;
Int I;
Queue <node> q;
N. x = SX;
N. Y = sy;
N. Time = 0;
Hash [n. x] [N. Y] = 6; // perfect Processing
// N. Count = 6;
Q. Push (N );
While (! Q. Empty ())
{
N = Q. Front ();
/*
If (N. x = EI & N. Y = EJ & hash [n. x] [N. Y]> 0)
{
Printf ("% d/N", N. time );
Return 1;
}
*/
/*
If (N. Time> = num)
{
Break;
}
*/
Q. Pop ();
For (I = 0; I <4; I ++)
{
P. x = n. x + dir [I] [0];
P. Y = n. Y + dir [I] [1];
// P. Count = n. Count-1;
P. Time = n. Time + 1;
If (hash [n. x] [n. y] <= 1 | P. x <0 | P. x> = n | P. Y <0 | P. y> = M | map [p. x] [p. y] = '0 ')
Continue;
/*
If (Map [p. x] [P. Y] = '1' | map [p. x] [P. Y] = '3 ')
{
Q. Push (P );
If (hash [p. x] [P. Y] Hash [p. x] [P. Y] = hash [n. x] [N. Y]-1;

}
*/
If (Map [p. x] [P. Y] = '3 ')
{
Printf ("% d/N", p. time );
Return 1;
}
If (hash [p. x] [P. Y] {
Hash [p. x] [P. Y] = hash [n. x] [N. Y]-1;
If (Map [p. x] [P. Y] = '4 ')
{
// P. Count = 6;
Hash [p. x] [P. Y] = 6;
// Q. Push (P );
}
Q. Push (P );
}

}

}
Printf ("-1/N ");
Return 1;

}

Int main ()
{
Int T;
Int I, J;
Cin> T;
While (t --)
{
Cin> N> m;
For (I = 0; I <n; I ++)
For (j = 0; j <m; j ++)
{
Cin> map [I] [J];
// Hash [I] [J] = 6;
If (Map [I] [J] = '2 ')
{
SI = I;
SJ = J;
}
If (Map [I] [J] = '3') // useless
{
Ei = I;
J = J;
}
}
// Num = 2 * (INT) (ABS (1.0 * si-EI) + ABS (1.0 * Sj-EJ ));
Memset (hash,-1, sizeof (hash ));
BFS (Si, SJ );
}
}