Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6461 Accepted Submission(s): 4087
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.
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)
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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
Source
Recommend
Eddy
code:
#include<iostream>
#include<cstring>
using namespace std;
char map[1005][1005];
int vis[1005][1005];
int sx,sy;
int ans;
int dir[4][2]={0,1,0,-1,1,0,-1,0};
int n,m;
void dfs(int sx,int sy)
{
if (map[sx][sy]=='.'||map[sx][sy]=='@')
{
ans++;
}
else
return ;
for (int i=0;i<4;i++)
{
int nowx=dir[i][0]+sx,nowy=dir[i][1]+sy;
if (nowx>=0&&nowy>=0&&nowy<m&&nowx<n&&map[nowx][nowy]=='.'&&vis[nowx][nowy]!=1)
{
// cout<<nowx<<" "<<nowy<<endl;
vis[nowx][nowy]=1;
dfs(nowx,nowy);
// vis[nowx][nowy]=0;
}
}
}
int main()
{
int sum;
int flag;
while (cin>>m>>n)
{
if(n==0&&m==0) break;
sum=0;
flag=0;
memset(map,0,sizeof(map));
memset(vis,0,sizeof(vis));
for (int i=0;i<n;i++)
for (int j=0;j<m;j++)
{
cin>>map[i][j];
vis[i][j]=0;
if (map[i][j]=='@')
{
flag=1;
sx=i;
sy=j;
}
}
// cout<<sx<<" "<<sy<<endl;
if (flag==1)
{
ans=0;
dfs(sx,sy);
cout<<ans<<endl;
}
else
cout<<0<<endl;
}
return 0;
}