Red and Black
Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 12519 Accepted Submission (s): 7753
Problem Descriptionthere is a rectangular and covered with square tiles. Each tile is colored either red or black. A man was standing on a black tile. From a tiles, he can move to one of the four adjacent tiles. But he can ' t move on red tiles, and 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.
Inputthe input consists of multiple data sets. A data set starts with a line containing the positive integers W and H; W and H is the numbers of tiles in the X-and y-directions, respectively. W and H is not more than 20.
There is H more lines in the data set, and each of the which includes W characters. 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)
Outputfor each data set, your program should output a line which contains the number of tiles he can reach from the Initia L tile (including itself).
Sample Input6 9 ..... .....# ...... ...... ...... ...... ...... #@...# .#.. #. 11 9. # ..... #.#######. .#.#.....#. .#.#.###.#. . #.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6.. #.. #.. #.. .. #.. #.. #.. .. #.. #.. ### .. #.. #.. #@. .. #.. #.. #.. .. #.. #.. #.. 7 7.. #.#.. .. #.#.. ###.### [email protected] ###.###. #.#.. .. #.#.. 0 0
Sample OUTPUT45 59 6 13 Simple DFS, no pruning; code:#include <stdio.h>
CharMap
[ +
][ +
];
int Dir
[4
][2
]={1
,0
,-1
,0
,0
,1
,0
,-1
},Step
,W
,H
;
void Dfs
(
intX
,
intY
){
int Tx
,Ty
;
Step
++;
for
(
intI
=0
;I
<4
;++I
){
Tx
=X
+Dir
[I
][0
];Ty
=Y
+Dir
[I
][1
];
If
(Tx
>=W
||Ty
>=H
||Tx
<0
||Ty
<0
||Map
[Ty
][Tx
]==' # '
||Map
[Ty
][Tx
]==' @ '
)
Continue
;
Map
[Ty
][Tx
]=' @ '
;
Dfs
(Tx
,Ty
);
}
}
int
Main
(){
intX
,Y
;
while
(scanf
("%d%d"
,&W
,&H
),W
||H
){Step
=0
;
for
(
intI
=0
;I
<H
;++I
)scanf
('%s '
,Map
[I
]);
for
(
intI
=0
;I
<H
;++I
)
for
(
intJ
=0
;J
<W
;++J
)
if
(Map
[I
][J
]==' @ '
)X
=J
,Y
=I
;
Dfs
(X
,Y
);
Printf
("%d\n"
,Step
);
}
return 0
;
}
Red and Black (Simple Dfs)
