今天做到了dfs的训练，感觉和bfs有相似之处，接下来用一道题来总结一下方法，可类比bfs。
上题：
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.
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)
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
解题代码：
#include<stdio.h>
#include<string.h>
char point[25][25];
bool flag[25][25]; //flag对应着我们需要研究的点point，用来标记是否曾经到过
int dx[4]={1,-1,0,0};
int dy[4]={0,0,-1,1};
int count;
void dfs(int x0,int y0,int r,int c)
{
for(int i=0;i<4;i++) //for循环用来探索所有邻接点
{
int tempx=x0+dx[i],tempy=y0+dy[i];
if(tempx<c&&tempx>=0&&tempy<r&&tempy>=0&&flag[tempy][tempx]==false&&point[tempy][tempx]=='.')
{
count++;
flag[tempy][tempx]=true; //满足条件且未标记的标记上
dfs(tempx,tempy,r,c); //通过递归来实现顺藤摸瓜的效果
}
}
return ;
}
void make_set()
{
for(int i=0;i<25;i++)
for(int j=0;j<25;j++)
flag[i][j]=false;
return ;
}
int main()
{
int c,r,x0,y0;
while(1)
{
count=1;
scanf("%d%d",&c,&r);
getchar();
if(c==0&&r==0)
break;
for(int i=0;i<r;i++)
{
for(int j=0;j<c;j++)
{
point[i][j]=getchar();
if(point[i][j]=='@')
{
x0=j;
y0=i;
}
}
getchar();
}
make_set();
dfs(x0,y0,r,c);
printf("%d\n",count);
}
return 0;
}
与bfs的区别：bfs是通过队列来进行逐层探索，而dfs则是沿着一个路径探索下去一直到底，到底后再返回沿其他路径探索，就和顺藤摸瓜差不多，大体上两者达到的效果基本一致（特殊情况效果有区别），两者均有缺点，bfs在编写代码时较麻烦，用到队列，一般要使用结构体，而dfs则是由于其使用递归调用，大数据易超时。
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赶紧跑回来加一句：最短路径用bfs！！