HDOJ 1312 DFS&BFS

Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7337 Accepted Submission(s): 4591

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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)

[align=left]Output[/align]
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).

[align=left]Sample Input[/align]

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.

11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........

11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..

7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

[align=left]Sample Output[/align]

45 59 6 13

广度搜索，需要用到队列：

#include<stdio.h>
int map[30][30];
int t,n,m;
int sx[4]={0,0,-1,1};
int sy[4]={-1,1,0,0};
void dfs(int h,int l)//深度搜索
{
int i,hx,hy;
t++;
map[h][l]='@';
for(i=0;i<4;i++)
{
hx=h+sx[i];
hy=l+sy[i];
if(hx>=1&&hx<=m&&hy>=1&&hy<=n&&map[hx][hy]=='.')
dfs(hx,hy);
}
}
int main()
{
int a,b,x,y;
while(scanf("%d%d",&n,&m)!=EOF)
{
if(n==0&&m==0)break;
for(a=1;a<=m;a++)
{
getchar();
for(b=1;b<=n;b++)
{
scanf("%c",&map[a][b]);
if(map[a][b]=='@')
{
x=a;
y=b;
}
}
}
t=0;
dfs(x,y);
printf("%d\n",t);
}
return 0;
}

View Code