poj1979 Red and Black

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 35903 Accepted: 19444

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

45596

#include<iostream>#include<cstring>#include<cstdio>using namespace std;int a,b,n,m;char c[22][22];int dfs(int a,int b){    if(c[a][b]=='#'||a>=m||b>=n||a<0||b<0)        return 0;    else    {        c[a][b]='#';        return 1+dfs(a,b+1)+dfs(a,b-1)+dfs(a-1,b)+dfs(a+1,b);    }}int main(){    int i,j;    while(scanf("%d%d",&n,&m)&&n&&m)    {        for(i=0;i<m;i++)        {            getchar();            for(j=0;j<n;j++)            {                scanf("%c",&c[i][j]);                if(c[i][j]=='@')                {                    a=i;b=j;                }            }        }            n=dfs(a,b);            printf("%d\n",n);    }    return 0;}