HDU 1312【DFS】

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24708    Accepted Submission(s): 14942

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

题解：

#include<cstdio>
#include<cstring>
#include<queue>
#include<iostream>
#define INF 99999999
using namespace std;
const int maxn=205;
char s[maxn][maxn];
int vis[maxn][maxn];
int minx=INF;
int n,m;
int dir[4][2]={0,1,-1,0,0,-1,1,0};
int cnt=1;
void dfs(int x,int y){
    for(int i=0;i<4;i++){
        int tx=x+dir[i][0];
        int ty=y+dir[i][1];
        if(tx<0||tx>=m||ty<0||ty>=n)
            continue;
        if(!vis[tx][ty]&&s[tx][ty]!='#'){
            cnt++;
            vis[tx][ty]=1;
            dfs(tx,ty);
        }
    }
}
int main(){
    while(~scanf("%d %d",&n,&m)&&(n+m)){
        memset(vis,0,sizeof vis);
        cnt=1;
        for(int i=0;i<m;i++){
            scanf("%s",s[i]);
        }
        int sx,sy;
        int flag=0;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(s[i][j]=='@'){
                    sx=i,sy=j;
                    flag=1;
                    break;
                }
            }
            if(flag) break;
        }
        vis[sx][sy]=1;
        dfs(sx,sy);
        printf("%d\n",cnt);
    }
    return 0;
}