杭电-1312 Red and Black（DFS）

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 17542    Accepted Submission(s): 10665

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

 

Source

Asia 2004, Ehime (Japan), Japan Domestic 

这道题题意很简单，用深搜搜索就行了

AC代码：

#include<stdio.h>#include<string.h>int m,n;int ans;char a[1100][1100];int dfs(int x,int y){    if(x< 0 || x >= n|| y < 0 ||y>=m)          return 0;      if(a[x][y]=='#')          return 0;      else{      ans++;           a[x][y]='#';              dfs(x,y+1);    dfs(x+1,y);    dfs(x,y-1);    dfs(x-1,y);     }      return ans;}  int main(){int i,j;while(scanf("%d%d",&m,&n)!=EOF){ans=0;if(m==0||n==0)break;int fi,fj;for(i=0;i<n;i++){scanf("%s",a[i]);}for(i=0;i<n;i++){for(j=0;j<m;j++){if(a[i][j]=='@'){fi=i;fj=j;}}}printf("%d\n",dfs(fi,fj));}return 0;}