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
   
   
   
   

搜索题；

好像又是种子填充；

难度不高；

#include<iostream>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<stack>
#include<queue>
#include<iomanip>
#include<map>
#define pi 3.14159265358979323846
using namespace std;
struct point
{
    int x,y;
};
point start;
int W,H;
char maze[25][25];
int dx[]={1,-1,0,0};
int dy[]={0,0,1,-1};
int ans;
bool flag[25][25];
void dfs(int x,int y)
{
    flag[x][y]=1;
    point temp;
    for(int i=0;i<4;++i)
    {
        temp.x=x+dx[i];
        temp.y=y+dy[i];
        if(temp.x>=0&&temp.x<H&&temp.y>=0&&temp.y<W&&flag[temp.x][temp.y]==0&&maze[temp.x][temp.y]=='.')
        {
            ++ans;
            dfs(temp.x,temp.y);
        }
    }
}
int main()
{
    while(scanf("%d %d",&W,&H)!=EOF&&W&&H)
    {
        for(int i=0;i<H;++i)
            for(int j=0;j<W;++j)
        {
            scanf(" %c",&maze[i][j]);
            if(maze[i][j]=='@')
            {
                start.x=i;
                start.y=j;
            }
        }
        ans=1;
        memset(flag,0,sizeof(flag));
        dfs(start.x,start.y);
        printf("%d\n",ans);
    }
    return 0;
}