hdu 1312 Red and Black

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

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

 

题解： 求 “.”数，  注意  行和列  吧...     找了好久才找到错在哪里...

 

代码：

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <limits.h> 
#include <algorithm>
#include <iostream>
#include <ctype.h>
#include <iomanip>
#include <queue>
#include <map>
#include <stdlib.h>
using namespace std;

int ans,m,n;
char v[30][30];
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};

void dfs(int x,int y)
{
    ans++;
    v[x][y]='#';
    for(int k=0;k<4;k++){
        int px=x+dx[k]; 
        int py=y+dy[k]; 
        if(v[px][py]=='.' && py<m && px<n && px>=0 && py>=0){ 
            dfs(px,py);
        }
    }
}

int main()
{
    while(~scanf("%d%d",&m,&n)&&m!=0&&n!=0){
        int e,d,i,j;
        for(i=0;i<n;i++)
            scanf("%s",v[i]);
        for(i=0;i<n;i++){
            for(j=0;j<m;j++){
                if(v[i][j]=='@'){
                    e=i;
                    d=j;
                }
            }
        }
        ans=0;
        dfs(e,d); 
        printf("%d\n",ans);
    }
}