hdoj 1312 Red and Black (DFS)

2015-01-24 05:42:23 · 作者: · 浏览: 3

Red and Black

http://acm.hdu.edu.cn/showproblem.php?pid=1312
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10361 Accepted Submission(s): 6472


Problem Description 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 integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of 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)

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

Sample Input
6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

Sample Output
45
59
6
13

Source Asia 2004, Ehime (Japan), Japan Domestic
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#include

#include
#include
#include
using namespace std;
char map[25][25];
int visit[25][25];
int fx[4]={1,0,-1,0};
int fy[4]={0,1,0,-1};
int w,h,step;
void DFS(int i,int j)
{
int k;
for(k=0;k<4;k++)
{
int x=fx[k]+i;
int y=fy[k]+j;
if(x>=0&&y>=0&&x {
step++;
map[x][y]='#';
DFS(x,y);
}

}
}
int main()
{
while(scanf("%d%d",&w,&h),(w||h))
{
int i,j,k;
step=1;
for(i=0;i scanf("%s",map[i]);

f or(i=0;i for(j=0;j if(map[i][j]=='@')
DFS(i,j);

printf("%d\n",step);
}
return 0;
}