Red and Black

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

题意：红格子和黑格子。人只能走黑格子，不能越过红色格子，则人能走多少个黑格子。

#include<stdio.h>#include<string.h>#define M 300using namespace std;int dx[4]={0,-1,1,0};int dy[4]={-1,0,0,1};int n,m,x,y,ans;bool vis[M][M];//标记走过的路char map[M][M];//输入的字符串int F(int xx,int yy){    int i,j;    for(i=0;i<4;i++)    {        if(map[xx+dx[i]][yy+dy[i]]=='.'&&vis[xx+dx[i]][yy+dy[i]]==0&&xx+dx[i]>=0&&yy+dy[i]>=0&&xx+dx[i]<m&&yy+dy[i]<n)        {            vis[xx+dx[i]][yy+dy[i]]=1;            ans++;            F(xx+dx[i],yy+dy[i]);        }    }}int main(){    while(scanf("%d%d",&n,&m)&&(n||m))    {        memset(vis,0,sizeof(vis));        for(int i=0;i<m;i++)        {            getchar();            for(int j=0;j<n;j++)            {                scanf("%c",&map[i][j]);                if(map[i][j]=='@')                {                    x=i,y=j;                    vis[i][j]=1;                }            }        }        ans=0;        F(x,y);        printf("%d\n",ans+1);    }    return 0;       }