A

转自楚江枫
Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20903 Accepted Submission(s): 12726

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
45
59
6
13

裸 dfs || bfs

附上 dfs + bfs

AC代码：

DFS ：

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;const int MAX = 22;typedef long long LL;char s[MAX][MAX];int vis[MAX][MAX],n,m,ans;int fx[4] = {0,0,-1,1},fy[4] = {-1,1,0,0};void dfs(int x,int y){    vis[x][y] = 1,ans++;    for(int i = 0; i < 4; i++){        int xx = x + fx[i],yy = y + fy[i];        if(xx >= 0 && yy >= 0 && xx < n && yy < m && !vis[xx][yy] && s[xx][yy] == '.')            dfs(xx,yy);    }}int main(){    while(~scanf("%d %d",&m,&n),n){        int x,y;        memset(vis,0,sizeof(vis));        for(int i = 0; i < n; i++)            scanf("%s",s[i]);        for(int i = 0; i < n; i++)            for(int j = 0; j < m; j++)                if(s[i][j] == '@')                    x = i,y = j;        ans = 0;        dfs(x,y);        printf("%d\n",ans);    }    return 0;}

bfs

#include<cstdio>#include<queue>#include<cstring>#include<algorithm>using namespace std;const int MAX = 22;typedef long long LL;char s[MAX][MAX];int vis[MAX][MAX],n,m,ans;int fx[4] = {0,0,-1,1},fy[4] = {-1,1,0,0};struct node{    int x,y;};void bfs(int x,int y){    vis[x][y] = 1;    node o;    o.x = x,o.y = y;    queue <node> q;    q.push(o);    while(!q.empty()){        o = q.front();        ans++,q.pop();        for(int i = 0; i < 4; i++){            int xx = o.x + fx[i],yy = o.y + fy[i];            if(xx >= 0 && yy >= 0 && xx < n && yy < m && !vis[xx][yy] && s[xx][yy] == '.'){                node w;                vis[xx][yy] = 1;                w.x = xx,w.y = yy;                q.push(w);            }        }    }}int main(){    while(~scanf("%d %d",&m,&n),n){        int x,y;        memset(vis,0,sizeof(vis));        for(int i = 0; i < n; i++)            scanf("%s",s[i]);        for(int i = 0; i < n; i++)            for(int j = 0; j < m; j++)                if(s[i][j] == '@')                    x = i,y = j;        ans = 0;        bfs(x,y);        printf("%d\n",ans);    }    return 0;}

贴上自己的代码
dfs

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;int n,m,x,y,ans;char mapp[110][110];int xx[4]={-1,1,0,0};int yy[4]={0,0,-1,1};void dfs(int a,int b){    if(a<0||a>=n||b<0||b>=m||mapp[a][b]=='#')    {        return ;    }    ans++;    mapp[a][b]='#';    for(int i=0;i<4;i++)    {        int dx=a+xx[i];        int dy=b+yy[i];        dfs(dx,dy);    }}int main()  {      while(scanf("%d%d",&m,&n)!=EOF&&(n!=0||m!=0))    {        ans=0;        getchar();        memset(mapp,0,sizeof(mapp));        for(int i=0;i<n;i++)        {            for(int j=0;j<m;j++)            {                scanf("%c",&mapp[i][j]);                if(mapp[i][j]=='@')                {                    x=i;                    y=j;                }            }            getchar();        }        dfs(x,y);        printf("%d\n",ans);    }    return 0;  }