hdoj 1312 Red and Black

Red and Black

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 71   Accepted Submission(s) : 65

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
题意：求@点最多能走多少步！#是墙，不能走！
搜索即可：
dfs：
#include<stdio.h>
int cnt=0;
int n,m;
char a[22][22];
void dfs(int x,int y)
{
    if(a[x][y]=='#')
    return ;
    if(y>=1&&y<=n&&x<=m&&x>=1)
    {
        a[x][y]='#';
        cnt++; 
    dfs(x+1,y);//四个方向搜索 
    dfs(x-1,y);
    dfs(x,y+1);
    dfs(x,y-1);
   }
}
int main()
{
    int i,j;
    while(scanf("%d%d",&n,&m),n|m)
    {
        cnt=0;
        int x,y;
            for(i=1;i<=m;i++)
            {
            getchar();
              for(j=1;j<=n;j++)
                {
                    scanf("%c",&a[i][j]);
                    if(a[i][j]=='@')//确定起点 
                    {
                        x=i;
                        y=j;
                    }
                }
            }
            dfs(x,y);
      printf("%d\n",cnt);
    }
}