Red and Black
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 44130 | Accepted: 23912 |
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)
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 13
题目大意：@表示人 . 表示黑瓷砖 #表示墙，人刚开始站在黑瓷砖上，问：人随便走可以最多可以走过多少瓷砖。
题解：很容易就想到用DFS，因为人是可以随便走的，所以只要用DFS搜索整个图，然后看递归多少次就是答案。
具体代码如下：
#include<iostream>
#include<cstring>
using namespace std;
const int N=21;
char map[N][N];
int m,n,ans;
int d[4][2]={0,1,1,0,0,-1,-1,0};
void DFS(int x,int y)
{
ans++;
map[x][y]='#';
for(int i=0;i<4;i++)
{
int nx=x+d[i][0];
int ny=y+d[i][1];
if(nx<0||ny<0||nx>=n||ny>=m) continue;
if(map[nx][ny]=='#') continue;
DFS(nx,ny);
}
}
int main()
{
while(cin>>m>>n&&m+n)
{
memset(map,'#',sizeof(map));
for(int i=0;i<n;i++)
cin>>map[i];
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
if(map[i][j]=='@')
{
ans=0;
DFS(i,j);
break;
}
cout<<ans<<endl;
}
return 0;
}