Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13464 Accepted Submission(s): 8338
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
Source
Asia 2004, Ehime (Japan), Japan Domestic
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题目分析：
最简单的dfs 就是让你 从@出发最对能走多少步 可以四个方向走但是遇到#不能走
#include<cstdio>
#include<cstring>
char a[22][22];
int x,y;
int n,m;
int cnt;
void dfs(int x,int y)
{
if(a[x][y]=='#') return;
if(x>=1&&x<=m&&y>=1&&y<=n)
{
a[x][y]='#';// 把走过的路都标记成墙不能再走了
cnt++;// 每走一步就加一
dfs(x+1,y);// 向下遍历
dfs(x-1,y);// 向上遍历
dfs(x,y-1);// 向左遍历
dfs(x,y+1);// 向右遍历 就是每个方向都走一遍直到遇到 # 转方向
}
}
int main()
{
while(~scanf("%d%d",&n,&m)&&n|m)
{
int i,j;
memset(a,0,sizeof(a));
for(i=1;i<=m;i++)
{
scanf("%s",a[i]+1);// 记住这里一定要加一因为 你的初始位置是 1 1 如果不加仪的话初始位置就 //是1 0了 因为这里选择的是一行一行的输入每行是一个字符串 字符串的首地址就是 0 所以这里要加1 变成 1 1
for(j=1;j<=n;j++)
{
if(a[i][j]=='@')
{
x=i;
y=j;
}
}
}
cnt=0;
dfs(x,y);
printf("%d\n",cnt);
}
return 0;
}