Red and Black
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 21482 | Accepted: 11488 |
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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
Source
思路 ：简单的深搜
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
int r,c;
char a[200][200];
int vis[200][200];
int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int sum;
void dfs(int i,int j)
{
//if(a[i][j]=='#'||(vis[i][j])||(i<0||j<0||(i>r-1||j>c-1)))
//return;
//else
//{
a[i][j]='#';
sum++;
for(int k = 0;k < 4;k++)
{
int x = i+dir[k][0];
int y = j+dir[k][1];
if(a[x][y]=='.'&&!vis[x][y]&&x>=0&&y>=0&&x<=r-1&&y<=c-1)
{
dfs(x,y);
vis[x][y]=1;
}
}
//}
}
int main()
{
int i,j,x,y;
while(scanf("%d%d",&c,&r),r||c)
{
sum=0;
memset(vis,0,sizeof(vis));
memset(a,0,sizeof(a));
for(i=0;i<r;i++)
{
getchar();
for(j=0;j<c;j++)
{
a[i][j] = getchar();
if(a[i][j]=='@')
{
x=i;
y=j;
}
}
}
//count=0;
dfs(x,y);
cout<<sum<<endl;
}
}