poj1979



Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 23928 Accepted: 12935

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)
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).

在一个空房间李，只能走黑色的砖，不能走红色的砖，走到无路可走就结束，不能重复走。

Sample Input

6 9....#......#..............................#@...#.#..#.11 9.#..........#.#######..#.#.....#..#.#.###.#..#.#..@#.#..#.#####.#..#.......#..#########............11 6..#..#..#....#..#..#....#..#..###..#..#..#@...#..#..#....#..#..#..7 7

..#.#....#.#..###.###...@...###.###..#.#....#.#..0 0
Sample Output
4559613
代码：
#include<stdio.h>char a[100][100];int map[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};int n, m;int num;void dfs(int x,int y){    num ++;    int xx, yy, k;    a[x][y] = '#';//把走过的路径堵死，不能返回。    for(k = 0; k < 4; k ++)    {        xx = x + map[k][0];        yy = y + map[k][1];        if(xx >= 0 && xx < m && yy >= 0 && yy < n && a[xx][yy] == '.')        {            dfs(xx,yy);        }    }}int main(){    int i, j, si, sj;    while(scanf("%d%d",&n,&m) != EOF)    {        if(n == 0 && m == 0)            break;        for(i = 0; i < m; i ++)        {            for(j = 0; j < n; j ++)            {                scanf(" %c",&a[i][j]);                if(a[i][j] == '@') //从@点开始走。                {                    si = i;//@点的坐标。                    sj = j;                }            }        }        num = 0;        dfs(si,sj);        printf("%d\n",num);    }    return 0;}

看了几遍题目，也没偶看懂，最后问了一下同学才知道题目的意思。才开始学习DFS，用晕的不是很熟练，参考着同学的代码才写出来的。