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最新推荐文章于 2024-07-06 00:20:02 发布

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本文介绍了一个关于深度优先搜索的应用实例——在一个黑白格子组成的矩形房间中，从指定的黑色格子出发，计算能够到达的所有黑色格子的数量。文章详细解释了如何使用递归方法实现深度优先搜索，并附上了完整的C++代码。

2107: Red and Black

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	375	207	Standard

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

This problem is used for contest: 17 150

Submit / Problem List / Status / Discuss

这个题是一个深度优先搜索的一个题，

从起始位置开始向四个方向进行搜索，

与其他深度优先搜索不同的是该题没有用到回溯。

只是对于可以到达的地方进行标记即可。

#include<cstdio>

int row,col;
int sum;
char map[1000][24];
void find(int &u,int &v)
{
for(int i=0;i<row;i++)
{
for(int j=0;j<col;j++)
{
if(map[i][j]=='@')
{
u=i,v=j;
return ;
}
}
}
}
void dfs(int u,int v)
{
//向上
if(u>0&&map[u-1][v]!='@'&&map[u-1][v]!='#')
{
map[u-1][v]='@';
sum++;
dfs(u-1,v);
}
//向下
if(u<row-1&&map[u+1][v]!='@'&&map[u+1][v]!='#')
{
map[u+1][v]='@';
sum++;
dfs(u+1,v);
}
//向左
if(v>0&&map[u][v-1]!='#'&&map[u][v-1]!='@')
{
map[u][v-1]='@';
sum++;
dfs(u,v-1);
}
//向右
if(v<col-1&&map[u][v+1]!='@'&&map[u][v+1]!='#')
{
map[u][v+1]='@';
sum++;
dfs(u,v+1);
}
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
while(scanf("%d%d",&col,&row),col&&row)
{
getchar();
for(int i=0;i<row;i++)gets(map[i]);
int u,v;
find(u,v);
sum=1;
dfs(u,v);
printf("%d\n",sum);
}
return 0;
}