杭电oj red and black（简单的深搜）

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Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12640 Accepted Submission(s): 7801

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
题意 找到@能经过的“.”的个数。
代码：
/*时间 2015/7/22 23:09 感受：收获甚大。 
*/ #include<stdio.h>
#include<string.h>#include<iostream>#include<algorithm>using namespace std;int dk[25][25];char dp[25][25];int ds[4][2]={{0,-1},{0,1},{-1,0},{1,0}};int W ,H;void dfs(int x,int y)//对数据进行搜索 { int i; for(i=0;i<4;i++) {  int dx=x+ds[i][0];  int dy=y+ds[i][1];  if(dx<H&&dx>=0&&dy>=0&&dy<W)  {   if(!dk[dx][dy]&&dp[dx][dy]=='.')   {    dk[dx][dy]=1;//把可行数据进行标记。     dfs(dx,dy);//继续搜索    }  } }}int main(){ int i,j; while(scanf("%d%d",&W,&H)!=EOF,W&&H) {  memset(dk,0,sizeof(dk));  for(i=0;i<H;i++)  scanf("%s",dp[i]);  int x,y;  for(i=0;i<H;i++)  for(j=0;j<W;j++)  if(dp[i][j]=='@')  {  x=i,y=j;  break;     }  dk[x][y]=1;  dfs(x,y);      int ans=0;//对标记数据进行统计。   for(i=0;i<H;i++)     for(j=0;j<W;j++)        if(dk[i][j]==1)          ans++;          printf("%d\n",ans); } return 0;}

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