POJ-1979

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 36371		Accepted: 19731

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) 
The end of the input is indicated by a line consisting of two zeros. 

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

题意：

@为起点，只能走' . '，求最多能走多少。

dfs求出所有可行路径，每走过一次则将' . '转换为' # '，防止重复。

AC代码：

#include<bits/stdc++.h>//注意poj提交时需要更换头文件
using namespace std;

int dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};
char mp[50][50];
int w,h,sx,sy;
int res;

void dfs(int x,int y){
    //cout<<mp[x][y]<<endl;
    if(mp[x][y]=='.'){
        res++;
        mp[x][y]='#';
        //cout<<x<<' '<<y<<endl;
    }
    else
    return ;
    for(int i=0;i<4;i++){
        x+=dx[i];
        y+=dy[i];
        //cout<<x<<" "<<y<<endl;
        dfs(x,y);
        x-=dx[i];
        y-=dy[i]; 
    }
    return ;
}

int main(){
    ios::sync_with_stdio(false);
    while(cin>>w>>h&&w&&h){
        memset(mp,'#',sizeof(mp));
        for(int i=0;i<h;i++){
            for(int j=0;j<w;j++){
                cin>>mp[i][j];
                if(mp[i][j]=='@'){
                    sx=i;
                    sy=j;
                }
            }
        }
        mp[sx][sy]='.';
        res=0;
        dfs(sx,sy);
        cout<<res<<endl;
    }
    return 0;
}