深搜水题 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).

6 9
….#.
…..#
……
……
……
……
……

@…

.#..#.
11 9
.#………
.#.#######.
.#.#…..#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#…….#.
.#########.
………..
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
0 0

Sample Output
45
59
13

#include<cstdio>#include<queue>#include<cstring>using namespace std;typedef struct A{    int x;    int y;}z;##BFSint main(){    int m,n,sum,i1,j1;    char a[22][22];    int f[4][2]={{1,0},{0,-1},{-1,0},{0,1}};    queue<z> s;    while(~scanf("%d%d",&n,&m))    {        if(m == 0 && n == 0)        {            break;        }        getchar();        memset(a,0,sizeof(a));        for(int i = 1;i <= m;++i)        {            for(int j = 1;j <= n;++j)            {                scanf("%c",&a[i][j]);                if(a[i][j] == '@')                {                    z temp;                    temp.x = i;                    temp.y = j;                    a[i][j] = '#';                    s.push(temp);                }            }            getchar();        }        sum = 1;        while(!s.empty())        {            for(int k=0;k<=3;++k)            {                z temp;                i1=s.front().x+f[k][0];                j1=s.front().y+f[k][1];                if(a[i1][j1]!=0 && a[i1][j1] == '.')                {                    a[i1][j1] = '#';                    sum++;                    temp.x = i1;                    temp.y = j1;                    s.push(temp);                }            }            s.pop();        }        printf("%d\n",sum);    }    return 0;}

#include<cstdio>#include<cstring>using namespace std;int dir[4][2] = {{1,0},{0,-1},{-1,0},{0,1}};int sum;int zong,heng;char obj[22][22];void dfs(int i,int j){    int x,y;    sum++;    obj[i][j] = '#';    for(int k = 0;k <= 3;++k)    {        x=i+dir[k][0];        y=j+dir[k][1];        if(obj[x][y]!=0 && obj[x][y]!='#')            dfs(x,y);    }}int main(){    while(~scanf("%d %d",&zong,&heng))    {        getchar();        int i1,j1;        if(zong == 0 && heng == 0)            break;        memset(obj,0,sizeof(obj));        for(int i = 1;i <= heng;++i)        {            for(int j = 1;j<=zong;++j)            {                scanf("%c",&obj[i][j]);                if(obj[i][j] == '@')                {                    i1 = i;                    j1 = j;                }            }            getchar();        }        sum = 0;        dfs(i1,j1);        printf("%d\n",sum);    }    return 0;}

总结:将顶点的所有邻接点遍历，邻接关系为上左下右。