HDU 1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12972    Accepted Submission(s): 8019

Problem 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

4559613

 

/**这是一个比较基础的搜索题，然而却用了我很多时间，费了很大的功夫才调试对，刚学搜索，用的还不熟练，这个题相当于一笔完成的那种趣味数学题，深搜就可以了*/#include<stdio.h>#include<string.h>int count = 0;int map[25][25];///地图int dir[4][2] = {{0,-1},{1,0},{0,1},{-1,0}};///定义四个方向int visited[25][25];///用于标记走过的点   初始化为0，走过的标志为1，int w, h, si, sj;///宽度，和高度，以及起点void dfs(int x, int y){    visited[x][y] = 1;    for(int i = 0; i < 4; i++)    {        int new_x = x + dir[i][0];        int new_y = y + dir[i][1];        /**其实不需要复杂的剪枝，刚开始的时候，写了控制边界和特殊点的控制条件        以至于看起来很麻烦，后来一想，其实完全没有必要，满足条件的进行判断，不满足的        不判断就好，地图初始化一下就可以了，超出边界的坑定不满足条件了*/        if(map[new_x][new_y] == '.' && !visited[new_x][new_y])        {            count++;///记录走过的点的个数            visited[new_x][new_y] = 1;///标记走过的点            dfs(new_x,new_y);///对新节点继续进行搜索        }    }}int main(){    char temp;    while(~scanf("%d%d",&w,&h)&&(w+h))    {        memset(map,'\0',sizeof(map));        scanf("%c",&temp);        for(int i = 1; i <= h; i++)        {            for(int j = 1; j <= w; j++)            {                scanf("%c",&map[i][j]);                if(map[i][j] == '@')                {                    si = i;                    sj = j;                }            }            scanf("%c",&temp);        }        count = 0;        memset(visited,0,sizeof(visited));///对标记数组初始化        dfs(si,sj);        printf("%d\n",count+1);    }    return 0;}