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 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

题目大意：

给出两个数m和n，代表n行m列，都不超过20，然后是n行m列的图，包括'.' , '#' , '@'3个字符。

@代表初始位置，'.'代表通路，‘#’代表墙，求最大可到达的 '.' 的数量 '@'算一个

BFS：

即求最大连通块问题 不一定非要用队列 一维数组一样可以解决

#include<stdio.h>
#include<string.h>
struct node
{
   int x,y;
}q[410];

struct node P, N;
int flag[25][25];
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
char str[25][25];

int main()
{
     int c, r, i, j, front, rear;
     while(scanf("%d%d",&c,&r)!=EOF, c + r){
        memset(flag, 0, sizeof(flag));
        for(i = 0; i < r; i++)
            scanf("%s", str[i]);

        for(i = 0; i < r; i++){
            for(j=0;j<c;j++)
                if(str[i][j] == '@') break;
            if(str[i][j] == '@') break;
        }
        N.x = i;
        N.y = j;
        flag[i][j] = 1;
        q[0] = N;
        front = 0;
        rear = 1;

        while(front < rear){
            N = q[front++];
            for(i = 0; i < 4; i++){
                int tx = N.x + dir[i][0];
                int ty = N.y + dir[i][1];
                if(tx >= 0 && tx < r && ty >= 0&& ty < c && flag[tx][ty] == 0 && str[tx][ty] == '.'){
                    P.x = tx;
                    P.y = ty;
                    q[rear++] = P;
                    flag[tx][ty] = 1;
                }
            }
        }
        printf("%d
",rear);
     }
     return 0;
}