hdu 1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2719    Accepted Submission(s): 1794

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
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    45
59
6
13
   
   
   
   

 

本题用深搜来做，简单！！！！

#include<iostream>

using namespace std;
char g[22][22];
int vis[22][22],count,xx,yy,w,h,sx,sy;
int dir[][2]={{1,0},{-1,0},{0,1},{0,-1}};
void dfs(int x,int y)
{
int i;
    if(g[y][x]=='#')return;
else
{
vis[y][x]=1;
for( i=0; i<4; i++ )
{
yy=y+dir[i][0];
xx=x+dir[i][1];
if( x<w && xx>=0 && y<h && yy>=0 && vis[yy][xx]==0)
{
dfs(xx,yy);
}
}
}
}
int main()
{
void dfs(int x,int y);
int i,j;
while( cin>>w>>h,w,h)
{
count=0;
for( i=0; i<h; i++)
{
for( j=0; j<w; j++)
{
cin>>g[i][j];
vis[i][j]=0;
}
}

for( i=0; i<h; i++)
{
for( j=0; j<w; j++)
{
if( g[i][j]=='@')
{
sx=j;
sy=i;

}
}
}
dfs(sx,sy);
//cout<<count<<endl;
        for( i=0; i<h; i++)
{
for( j=0; j<w; j++)
{
  if( vis[i][j]==1)
  {
  count++;
  }
}
}
cout<<count<<endl;
}
return 0;
}