Zoj 2165 red and black

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 integersWAndH;WAndHAre the numbers of tiles inX-AndY-Directions, respectively.WAndHAre not more than 20.

There areHMore lines in the data set, each of which limit desWCharacters. 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 shocould output a line which contains the number of tiles he can reach from the initial tile (including itself ).

Sample Input

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
. #. # [Email protected] #. #.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7
..#.#..
..#.#..
###.###
[Email protected]
###.###
..#.#..
..#.#..
0 0

Sample output

45
59
6
13

 

Question:

Two numbers m and n are given, representing n rows of M columns, which cannot exceed 20. Then, the diagram of N rows of M columns includes '. ',' # ',' @ ', 3 characters.

@ Indicates the initial position, '.' indicates the path, '#' indicates the wall, and calculates the maximum number of '.' to be reached. '@' is

 

BFS:

That is to say, the maximum connected block problem does not have to be solved using the same one-dimensional array of the queue.

 1 #include<stdio.h> 2 #include<string.h> 3 struct node 4 { 5    int x,y; 6 }q[410]; 7  8 struct node P, N; 9 int flag[25][25];10 int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};11 char str[25][25];12 13 int main()14 {15      int c, r, i, j, front, rear;16      while(scanf("%d%d",&c,&r)!=EOF, c + r){17         memset(flag, 0, sizeof(flag));18         for(i = 0; i < r; i++)19             scanf("%s", str[i]);20 21         for(i = 0; i < r; i++){22             for(j=0;j<c;j++)23                 if(str[i][j] == ‘@‘) break;24             if(str[i][j] == ‘@‘) break;25         }26         N.x = i;27         N.y = j;28         flag[i][j] = 1;29         q[0] = N;30         front = 0;31         rear = 1;32 33         while(front < rear){34             N = q[front++];35             for(i = 0; i < 4; i++){36                 int tx = N.x + dir[i][0];37                 int ty = N.y + dir[i][1];38                 if(tx >= 0 && tx < r && ty >= 0&& ty < c && flag[tx][ty] == 0 && str[tx][ty] == ‘.‘){39                     P.x = tx;40                     P.y = ty;41                     q[rear++] = P;42                     flag[tx][ty] = 1;43                 }44             }45         }46         printf("%d\n",rear);47      }48      return 0;49 }