红与黑（dfs水题）

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本文介绍了一个迷宫寻路问题的解决方法，通过深度优先搜索算法遍历地图，找到所有可达的黑色方块数量。该算法将起点标记为不可通行，并递归地探索四个相邻的方块。

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Red and Black

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

解题思路：这题主要是讲的一种做题方法，就是'@'经过之后可以直接转化为'#',然后处理起来就会方便很多的，首先统计字符'.'的个数，然后减去经过处理之后的'.'的个数，再加上一个@的数值，就是能够变色的大小。

#include <stdio.h> #include <string.h> int n,m,p,q; char map[25][25]; void dfs(int x,int y) { int tx,ty,k; int next[4][2]={{1,0},{0,1},{-1,0},{0,-1}}; for(k=0;k<4;k++) { tx=x+next[k][0]; ty=y+next[k][1]; if(tx<1||tx>n||ty<1||ty>m) continue; if(map[tx][ty]=='.') { map[tx][ty]='#'; dfs(tx,ty); } } return ; } int main() { int i,j,sum,s,h; while(scanf("%d%d",&m,&n),n!=0,m!=0) { getchar(); s=sum=0; for(i=1;i<=n;i++) { scanf("%s",map[i]+1); for(j=1;j<=m;j++) { if(map[i][j]=='@') { p=i; q=j; } if(map[i][j]=='.') sum++; if(map[i][j]=='#') h++; } } map[p][q]='#'; dfs(p,q); for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { if(map[i][j]=='.') s++; //printf("%c",map[i][j]); } //printf("\n"); } if(h==0) printf("%d\n",n*m); else printf("%d\n",sum-s+1); } return 0; }