POJ-1979 Red and Black

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 35471		Accepted: 19195

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

Source
Japan 2004 Domestic
题意：从符号'@'出发能踩多少个黑块'.'（不能站在红块'#'上），出发点也算一个黑块。
思路：DFS，从出发点朝四个方向走，只要没有碰到红块且坐标合法，可以踩到的黑块数+1，并把黑块变成红块，避免下一次再访问，递归调用DFS访问接下来的节点，直到访问完成或者四面碰壁，结束退出，输出结果。

#include<iostream>
#include<cstring>
using namespace std;
char graph[20][20]={0};
int i,j,p,q;
int I,J;
int result=0;
void DFS(int p,int q)
{
if(p>=0 && p<=I-1 && q>=0 && q<=J-1 && graph[p][q]=='.')
{
result++;
graph[p][q]='#';
}
else   return ;
DFS(p-1,q);
DFS(p,q-1);
DFS(p,q+1);
DFS(p+1,q);
}
int main()
{
while((cin >> J >> I) && I && J)
{
result=0;
memset(graph,0,sizeof(graph));
for(i=0;i<I;i++)
{
for(j=0;j<J;j++)
{
cin >> graph[i][j];
if(graph[i][j]=='@')
{
p=i;q=j;
graph[i][j]='.';
}
}
}
DFS(p,q);
cout << result << endl;
}
return 0;
}