J - Red and Black 寒假练习1-J

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

来源：zoj2165

解题思路：

找到格点‘@’，以‘#’为基础广搜，输出找到的‘.’,格点个数。

代码：

#include<stdio.h>
#include<string.h>
char map[105][105];
int m,n,v;
int next[4][2]={1,0,0,1,-1,0,0,-1};
void bfs(int x,int y)
{
for(int k=0;k<4;k++)
{

int tx=x+next[k][0];
int ty=y+next[k][1];

if(map[tx][ty]!='.'||tx<0||tx>=n||ty<0||ty>=m)
continue;
//      printf("%d %d m=%d n=%d\n",tx,ty,m,n);
if(map[tx][ty]=='.')
{
v++;
map[tx][ty]='#';
bfs(tx,ty);
}
}
}
int main()
{
while(scanf("%d%d",&m,&n),m||n)
{
int i,j;
for(i=0;i<n;i++)
scanf("%s",map[i]);
int f=0;
//      for(i=0;i<n;i++)
//      {
//          for(j=0;j<m;j++)
//          {
//              printf("%c ",map[i][j]);
//          }
//          printf("\n");
//      }
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
if(map[i][j]=='@')
{
f=1;
break;
}
}
if(f==1)
break;
}
v=1;
map[i][j]='#';
bfs(i,j);
printf("%d\n",v);
}
return 0;
}

错误解析：初始点走过了就不再走了，搜索时注意边界问题。