POJ 1979 - Red and Black（深搜）

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 19423		Accepted: 10368

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) 
The end of the input is indicated by a line consisting of two zeros. 

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

===============================

就是一个简单的深搜

#include <iostream>
#include <cstdio>

using namespace std;

int n,m,ans,dir[4][2]={{-1,0},{0,-1},{0,1},{1,0}};
char map[22][22];

void dfs(int x,int y)
{
    map[x][y]='#';
    for(int i=0;i<4;i++)
    {
        int dx=x+dir[i][0];
        int dy=y+dir[i][1];
        if(map[dx][dy]=='.'&&dx>=1&&dx<=n&&dy>=1&&dy<=m)
        {
            ans++;
            dfs(dx,dy);
        }
    }
    return;
}

int main()
{
    int x,y;
    while(~scanf("%d%d",&m,&n))
    {
        if(n==0&&m==0) break;
        ans=1;
        for(int i=1;i<=n;i++)
            scanf("%s",map[i]+1);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(map[i][j]=='@')
                {
                    x=i;y=j;
                    break;
                }
            }
        }
        dfs(x,y);
        printf("%d\n",ans);
    }
    return 0;
}