poj1979 DFS
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)
The end of the input is indicated by a line consisting of two zeros.
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 13Source
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<map>
#include<algorithm>
#include<set>
#define INF 0x3f3f3f3f
using namespace std;
int n,m;
char a[300][300];
int vis[300][300];
int lx,ly;
int dx[]= {-1,1,0,0};
int dy[]= {0,0,-1,1};
int num=0;
int dp[300][300];
int panduan(int x,int y)
{
if(x>=0&&x<n&&y>=0&&y<m)
return 1;
return 0;
}
int DFS(int x,int y)
{
if(!panduan(x,y)||a[x][y]=='#')
return 0;
if(vis[x][y]==0)
{
vis[x][y]=1;
for(int i=0; i<4; i++)
{
int fx=dx[i]+x;
int fy=dy[i]+y;
if(panduan(fx,fy)&&vis[fx][fy]==0&&a[fx][fy]=='.')
{
DFS(fx,fy);
num++;
}
}
//vis[x][y]=0; //走过的点就不用再走了
}
return 1;
}
int main()
{
while(~scanf("%d%d",&m,&n))
{
if(!m&&!n)
break;
int flag=1;
memset(vis,0,sizeof(vis));
for(int i=0; i<n; i++)
{
scanf("%s",a[i]);
if(flag)
{
for(int j=0; j<m; j++)
{
if(a[i][j]=='@')
{
flag=0;<span id="transmark"></span> lx=i;ly=j;
break;
}
}
}
}
num=0;
DFS(lx,ly);
printf("%d\n",num+1);
}
}