hdu-1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19524    Accepted Submission(s): 11879

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

....#.

.....#

......

......

......

......

......

#@...#

.#..#. 

 

 

Sample Output

 

45

 

 

题意：

@为起点，可上下左右走，不能走#，问可以走多少.

 

简单DFS

 

附AC代码：

#include<bits/stdc++.h>
using namespace std;

int ans,sx,sy,n,m;
char ma[30][30];
int a[4][2]={1,0,-1,0,0,1,0,-1};

void dfs(int x,int y){
    ans++;
    ma[x][y]='#';
    for(int i=0;i<4;i++){
        int tx=x+a[i][0];
        int ty=y+a[i][1];
        if(tx<m&&tx>=0&&ty<n&&ty>=0&&ma[tx][ty]=='.'){
            dfs(tx,ty);
        }
    }
    return ;
}

int main(){
    while(cin>>n>>m){
        if(n==0&&m==0)
        break;
        ans=0;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                cin>>ma[i][j];
                if(ma[i][j]=='@'){
                    sx=i;
                    sy=j;
                    //cout<<i<<" "<<j<<endl;
                }
            }
        }
        dfs(sx,sy);
        cout<<ans<<endl;
    }
    return 0;
}