Poj 3984 Maze problem BFS

/*

Question:

Calculate the minimum time from (0, 0) to (4, 4) path

Analysis:

This is a pure BFS question. However, to print the path, you can use an array to record the previous coordinate of the current coordinate,

Because BFS constructs a BFS optimal Spanning Tree, each node's parent node is unique.

ReferenceAlgorithmIntroduction...

*/

# Include <iostream>

# Include <cstring>

# Include <cstdio>

# Include <queue>

Using namespace STD;

# Define x 6

Int map [x] [X], pre [x] [x];

Bool visit [x] [x];

Struct Node

{

Int X, Y;

};

Void print (int x, int y) // print the path

{

If (X | Y) // recursive until (0, 0) at the beginning, and then output continuously to the path (4, 4 ).

Print (pre [x] [Y]/10, pre [x] [Y] % 10 );

If (pre [x] [Y]! =-1 & Pre [x] [Y]! =-1) // do not output the coordinates before (0, 0...

Printf ("(% d, % d) \ n", pre [x] [Y]/10, pre [x] [Y] % 10 );

}

Void BFS ()

{

Queue <node> q;

Node temp, cur;

Temp. x = 0;

Temp. Y = 0;

Q. Push (temp );

Int X, Y;

Visit [0] [0] = true;

While (! Q. Empty () // BFS Core

{

Cur = Q. Front ();

Q. Pop ();

X = cur. X;

Y = cur. Y;

If (x = 4 & Y = 4)

Print (4, 4 );

If (X &&! Visit [x-1] [Y] &! Map [x-1] [Y]) // go up

{

Visit [x-1] [Y] = true;

Temp. x = X-1;

Temp. Y = y;

Pre [x-1] [Y] = x * 10 + Y;

Q. Push (temp );

}

If (Y &&! Map [x] [Y-1] &! Visit [x] [Y-1]) // left

{

Visit [x] [Y-1] = true;

Temp. x = X;

Temp. Y = Y-1;

Pre [x] [Y-1] = x * 10 + Y;

Q. Push (temp );

}

If (Y <4 &&! Map [x] [Y + 1] &! Visit [x] [Y + 1]) // go to the right

{

Visit [x] [Y + 1] = true;

Temp. x = X;

Temp. Y = Y + 1;

Pre [x] [Y + 1] = x * 10 + Y;

Q. Push (temp );

}

If (x <4 &&! Map [x + 1] [Y] &! Visit [x + 1] [Y]) // go down

{

Visit [x + 1] [Y] = true;

Temp. x = x + 1;

Temp. Y = y;

Pre [x + 1] [Y] = x * 10 + Y;

Q. Push (temp );

}

}

}

Int main ()

{

Freopen ("sum. In", "r", stdin );

Freopen ("sum. Out", "W", stdout );

While (CIN> map [0] [0])

{

Memset (PRE,-1, sizeof (pre ));

Memset (visit, false, sizeof (visit ));

For (INT I = 0; I <5; I ++)

For (Int J = 0; j <5; j ++)

If (I | j)

Scanf ("% d", & map [I] [J]);

BFS ();

Printf ("(% d, % d) \ n",); // print the final Coordinate

}

Return 0;

}