hdu1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 19354 Accepted Submission(s): 11772

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

思路：用到深搜，是其中一种比较简单的，分四个方向，标记一下，深搜出结果

#include<stdio.h>
#include<string.h>
#define N 22
char a

;
int vis

;
int w,h;
int move[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
int ans;
void dfs(int x,int y)
{
for(int i=0;i<4;i++)
{
int tx=x+move[i][0];
int ty=y+move[i][1];
if(tx>=0&&tx<h&&ty>=0&&ty<w&&a[tx][ty]=='.'&&!vis[tx][ty])
{
vis[tx][ty]=1;
ans++;
dfs(tx,ty);
}
}
}
int main()
{

while(scanf("%d%d",&w,&h)&&w+h)
{
ans=0;
for(int i=0;i<h;i++)
{
scanf("%s",a[i]);
}
memset(vis,0,sizeof(vis));
for(int i=0;i<h;i++)
{
for(int j=0;j<w;j++)
{
if(a[i][j]=='@')

{
vis[i][j]=1;
ans++;
dfs(i,j);
}
}
}
printf("%d\n",ans);
}
return 0;
}