hdu 1312 Red and Black (dfs)

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7328    Accepted Submission(s): 4585

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

 

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//0MS    256K    748 B    C++
/*

    题意：
        求与'@'连在一起的'.' 的数量
    
    DFS：
        基础dfs.. 

*/ 
#include<stdio.h>
#include<string.h>
char g[25][25];
int mov[4][2]={0,1,1,0,0,-1,-1,0};
int n,m,s,e;
int cnt;
void dfs(int x,int y)
{
    for(int i=0;i<4;i++){
        int tx=x+mov[i][0];
        int ty=y+mov[i][1];
        if(tx>=0 && tx<n && ty>=0 && ty<m && g[tx][ty]=='.'){
            cnt++;
            g[tx][ty]='#';
            dfs(tx,ty);
        }
    }
}
int main(void)
{
    while(scanf("%d%d",&m,&n),n+m)
    {
        for(int i=0;i<n;i++){
            scanf("%s",g[i]);
            for(int j=0;j<m;j++)
                if(g[i][j]=='@'){
                    s=i,e=j;
                }
        }
        cnt=1;
        g[s][e]='#';
        dfs(s,e);
        printf("%d
",cnt);
    }
    return 0;
}