Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12907 Accepted Submission(s): 7983
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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
#include<iostream>
#include<sstream>
#include<algorithm>
#include<cstdio>
#include<string.h>
#include<cctype>
#include<string>
#include<cmath>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
using namespace std;
const int INF=40;
char cnt[INF][INF];
int vis[INF][INF];
int dir[][2]= {{1,0},{0,1},{-1,0},{0,-1}};
int ans,w,h;
void dfs(int x,int y)
{
for(int i=0; i<4; i++)
{
int tx=dir[i][0]+x;
int ty=dir[i][1]+y;
if(tx>=1&&tx<=w&&ty>=1&&ty<=h&&!vis[tx][ty]&&cnt[tx][ty]=='.')
{
ans++;
vis[tx][ty]=1;
dfs(tx,ty);
}
}
return ;
}
int main()
{
while(cin>>h>>w,h+w)
{
memset(vis,0,sizeof(vis));
int si,sj;
ans=1;
for(int i=1; i<=w; i++)
scanf("%s",cnt[i]+1);
for(int i=1; i<=w; i++)
{
for(int j=1; j<=h; j++)
{
if(cnt[i][j]=='@')
{
si=i;
sj=j;
break;
}
}
}
dfs(si,sj);
cout<<ans<<endl;
}
return 0 ;
}
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