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). DFS&BFS
#include <stdio.h>
#include <string.h>
using namespace std;
#define MAXN 50
char str[50][50];
int vis[MAXN][MAXN];
int d[4][2]={0,1,0,-1,-1,0,1,0};
int n,m,ans=1;
void DFS(int x1,int y1)
{
vis[x1][y1]=1;
int x2,y2;
for(int i=0;i<4;i++)
{
x2=x1+d[i][0];y2=y1+d[i][1];
if(x2<0||x2>=m||y2<0||y2>=n) continue;
if(str[x2][y2]=='#') continue;
if(vis[x2][y2]==1) continue;
if(str[x2][y2]=='.')
{
ans++;
vis[x2][y2]=1;
DFS(x2,y2);
}
}
}
int main()
{
int stx,sty;
while(scanf("%d%d",&n,&m)&&(n+m))
{
for(int i=0;i<m;i++)
scanf("%s",str[i]);
for(int i=0;i<m;i++)
{
for(int j=0;j<n;j++)
if(str[i][j]=='@') stx=i,sty=j;
// DFS(stx,sty);
}
DFS(stx,sty);
printf("%d\n",ans);
memset(vis,0,sizeof(vis));
ans=1;
}
}
、、、、、、
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
#define MAXN 50
struct node{
int x,y,step;
bool friend operator < (node a,node b)
{
return a.step<b.step;
}
};
int n,m;
char str[MAXN][MAXN];
int vis[MAXN][MAXN];
int d[4][2]={0,1,0,-1,1,0,-1,0};
void BFS(int x1,int y1)
{
priority_queue<node>que;
node e1,e2;
memset(vis,0,sizeof(vis));
e1.x=x1,e1.y=y1,e1.step=1;
int ans=1;
que.push(e1);
vis[x1][y1]=1;
while(!que.empty())
{
e1=que.top();
que.pop();
for(int i=0;i<4;i++)
{
e2.x=e1.x +d[i][0];e2.y=e1.y+d[i][1];
if(e2.x<0||e2.x>=m||e2.y<0||e2.y>=n) continue;
if(str[e2.x][e2.y]=='#') continue;
if(vis[e2.x][e2.y]==1) continue;
if(str[e2.x][e2.y]=='.')
{
e2.step=e1.step+1;
que.push(e2);
vis[e2.x][e2.y]=1;
ans++;
}
}
}
printf("%d\n",ans);
}
int main()
{
int stx,sty;
while(scanf("%d%d",&n,&m)&&(n+m))
{
for(int i=0;i<m;i++) scanf("%s",str[i]);
for(int i=0;i<m;i++)
{
for(int j=0;j<n;j++)
{
if(str[i][j]=='@')
stx=i,sty=j;
}
}
BFS(stx,sty);
}
}Sample Input
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13