POJ 1979 Red and Black

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B - Red and Black

Time Limit:1000MS Memory Limit:30000KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1979

Appoint description: System Crawler (2018-02-12)

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

#include <iostream>
#include <climits>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <vector>

#define ll long long
#define REP(i, n) for (int i=0;i<n;++i)
#define REP_1(i, n) for (int i=1;i<=n;++i)
#define REP_2(i, j, n, m) REP(i, n) REP(j, m)
#define REP_2_1(i, j, n, m) REP_1(i, n) REP_1(j, m)

#define JU_RAN(a,x,b) a<=x&&x<=b



using namespace std;

int m,n;
char v[22][22];

int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};

int ans;
void dfs(int x,int y){
    ans++;
    v[x][y]='#';
    REP(i,4){
        int nx=x+dx[i];
        int ny=y+dy[i];
        if(JU_RAN(1,nx,n)&&JU_RAN(1,ny,m)&&v[nx][ny]=='.'){
            dfs(nx,ny);
        }
    }
}

void solve(){

    while (cin>>m>>n){
        if(m==0&&n==0){
            break;
        }
        int fx,fy;
        REP_2_1(i,j,n,m){
                cin>>v[i][j];
                if(v[i][j]=='@'){
                    fx=i;
                    fy=j;
                }
            }

        ans=0;
        dfs(fx,fy);
        cout<<ans<<endl;
    }








}



int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);

#ifdef LOCAL_DEFINE
    freopen("input.txt", "rt", stdin);
#endif

    solve();



#ifdef LOCAL_DEFINE
    cerr << "\nTime elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
    return 0;
}