hdu 1312

Red and Black

[b]Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2719 Accepted Submission(s): 1794

[/b]

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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)

[align=left]Output[/align]
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).

[align=left]Sample Input[/align]

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

[align=left]Sample Output[/align]

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;

}