http://acm.hdu.edu.cn/showproblem.php?pid=1312
Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6095 Accepted Submission(s): 3883
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.
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)
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
Source
dfs模板题
1 #include<stdio.h> 2 #include<string.h> 3 char map[25][25]; 4 int dir[4][2]={1,0,-1,0,0,1,0,-1};//方向数组 5 int vis[25][25];//标记数组 6 int n,m,count;//行，列，计数器 7 void dfs(int x,int y)//标准的DFS模板 8 { 9 int i; 10 for(i=0;i<4;i++) 11 { 12 int xx=x+dir[i][0]; 13 int yy=y+dir[i][1]; 14 if(xx>=0&&xx<n&&yy>=0&&yy<m&&!vis[xx][yy]&&map[xx][yy]!='#') 15 { 16 count++; 17 vis[xx][yy]=1; 18 dfs(xx,yy); 19 } 20 } 21 // return ; 22 } 23 int main() 24 { 25 int i,j; 26 while(scanf("%d %d",&m,&n)!=EOF) 27 { 28 if(m==0||n==0) 29 break; 30 memset(map,0,sizeof(map));//因为是多组输入所以每次得清0 31 memset(vis,0,sizeof(vis)); 32 count=1; 33 for(i=0;i<n;i++) 34 scanf("%s",map[i]); 35 for(i=0;i<n;i++) 36 for(j=0;j<m;j++) 37 { 38 if(map[i][j]=='@') 39 { 40 map[i][j]='#'; 41 vis[i][j]=1; 42 dfs(i,j); 43 } 44 } 45 printf("%d ",count); 46 } 47 return 0; 48 }