Solutions Using DFS in zoj 2165

Recently learning Graph TheoryAlgorithmIn fact, I have been learning for a long time and found that I have not made any progress. The reason is that I writeCodeThe time is still too short. Now, I start to use what I learned to solve some problems and help myself better understand the essence of algorithms! Come on!

In zoj, I found some questions and wrote them!

The following is a question!

 

There is a rectangular room, covered with square tiles. each tile is colored either red or black. A man is standing on a black tile. from a tile, he can move to one of four adjacent tiles. but he can't move on red tiles, he can move only on black tiles.

 

Write a program to count the number of black tiles which he can reach by repeating the moves described above.

 

Input

 

The input consists of multiple data sets. A data set starts with a line containing two positive integersWAndH;WAndHAre the numbers of tiles inX-AndY-Directions, respectively.WAndHAre not more than 20.

 

There areHMore lines in the data set, each of which limit desWCharacters. Each character represents the color of a tile as follows.

 

• '.'-A black Tile
• '#'-A Red Tile
• '@'-A man on a black tile (appears exactly once in a data set)

 

Output

 

For each data set, your program shocould output a line which contains the number of tiles he can reach from the initial tile (including itself ).

 

Sample Input

Sample Input

6 9

....#.

.....#

......

......

......

......

......

#@...#

.#..#.

11 9

.#.........

.#.#######.

.#.#.....#.

.#.#.###.#.

.#.#..@#.#.

.#.#####.#.

.#.......#.

.#########.

...... 11 6

..#..#..#..

..#..#..#..

..#..#..###

..#..#..#@.

..#..#..#..

..#..#..#..

7

..#.#..

..#.#..

###.###

...@...

###.###

..#.#..

..#.#..

0 0

Sample output

45

59

6

13

Below is the code I wrote using DFS search!

# Include <cstdio>
CharMap [21] [21];
IntM, N;
IntCount;
IntDir [4] [2] = {{0,1},{0,-1},{1,0},{-1,0}; // Indicates the direction of movement. This array can be used to represent
IntDFS (IntX,IntY)
{
IntI, XX, YY;
Map [x] [Y] ='#';
For(I =0; I <4; I ++)
{
Xx = x + dir [I] [0];
YY = Y + dir [I] [1];
If(Xx <0| YY <0| XX> = M | YY> = N)
Continue;
If(Map [XX] [YY] ='.')
{
Map [XX] [YY] ='#';
Count ++;
//Printf ("% d", count );
DFS (XX, YY );
}
}
ReturnCount;
}
IntMain ()
{
While(1)
{
IntI, j, X, Y;
Scanf ("% D", & N, & M );
If(M =0& Amp; n =0)
Break;
Char temp;
Scanf ("% C", & temp); // you have already entered a number on the current side. If you want to enter a number next to it, press Enter, and enter is a character, so it should be absorbed,This method can be used now!
For(I =0; I <m; I ++)
{
 For(J =0; J <n; j ++)
{
Scanf ("% C", & Map [I] [J]);
If(Map [I] [J] ='@')
{
X = I;
Y = J;
}
}
Scanf ("% C", & Temp );
}
Count =0;
DFS (x, y );
Printf ("% D \ n", Count +1);// Because @ is also a black brick, Count must be added!
}
}

This is to use DFS to calculate the number of bricks in a maze. Now I can only think of using DFS for search. In fact, the use of DFS is very simple, but it is not very easy to understand. Now I will summarize it!

When an image is stored using an adjacent matrix,

DFS (vertex I)

{Visited [I] = 1;

For (j = 0; j <n; j ++) // for all other vertex J

{

If (edge [I] [J] = 1 & visted [J]) // vertex J has not been accessed, and J is the adjacent vertex of I!

{

// Write the code before recursive search

 

DFS (vertex J)

// The following is the rollback position, but you need to write code in this place to restore the original situation!

}

}

}

If an adjacent table is used to store a graph

DFS (vertex I)

{Vistited [I] = 1;

// P is the header pointer of I;

While (P! = NULL)

{

If (visit [J]! = 1)

{

DFS (j );

}

}

}

This is a summary of some things. In the actual process, we need to learn more!