Red and Black
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 25 Accepted Submission(s) : 20
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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
算法：
深度优先搜索（DFS）
思路：
从“@”出发，向与其相邻的上下左右走，走过并标记，当无路可走时，回到上一个节点，搜索下一个方向，直到全部搜索完毕。
代码：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
using namespace std;
int x,y,a[23][23],map[23][23],t;
int go(int i, int j)
{
int k;
if (map[i][j]==1)
{
t++;
map[i][j]=0;
if (a[i+1][j]==a[i][j]) go(i+1,j);
if (a[i-1][j]==a[i][j]) go(i-1,j);
if (a[i][j+1]==a[i][j]) go(i,j+1);
if (a[i][j-1]==a[i][j]) go(i,j-1);
}
return t;
}
int main ()
{
int i,j,w,h;
char ch;
while (cin>>y>>x)
{
if (x==0 && y==0) break;
t=0;
memset(a,0,sizeof(a));
memset(map,1,sizeof(map));
for (i=0; i<x; i++)
{
for (j=0; j<y; j++)
{
cin>>ch;
if (ch=='.')
a[i][j]=1;
else if (ch=='#')
a[i][j]=0;
else
{
a[i][j]=1;
w=i;
h=j;
}
map[i][j]=1;
}
}
go(w,h);
cout<<t<<endl;
}
return 0;
}