速刷一道DFS
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.
@是起点，#不能走，'.'是正常道路，问从起点能到达多少格子
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)
The end of the input is indicated by a line consisting of two zeros.
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)
The end of the input is indicated by a line consisting of two zeros.
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
刷道普通的DFS，复习模板
1 /*by SilverN*/ 2 #include<iostream> 3 #include<algorithm> 4 #include<cstring> 5 #include<cstdio> 6 #include<cmath> 7 using namespace std; 8 const int mxn=50; 9 char mp[mxn][mxn]; 10 int ans; 11 int W,H; 12 void dfs(int x,int y){ 13 if(x<1 || x>H || y<1 || y>W)return; 14 if(mp[x][y]=='#')return; 15 ans++; 16 mp[x][y]='#'; 17 dfs(x+1,y); 18 dfs(x,y+1); 19 dfs(x-1,y); 20 dfs(x,y-1); 21 return; 22 } 23 int main(){ 24 while(scanf("%d%d",&W,&H) && W && H){ 25 memset(mp,' ',sizeof(mp)); 26 ans=0; 27 int i,j; 28 for(i=1;i<=H;i++){ 29 scanf("%s",mp[i]+1); 30 } 31 int sx,sy; 32 bool flag=0; 33 for(i=1;i<=H;i++){ 34 for(j=1;j<=W;j++) 35 if(mp[i][j]=='@'){ 36 sx=i;sy=j; 37 flag=1; 38 break; 39 } 40 if(flag)break; 41 } 42 dfs(sx,sy); 43 printf("%d ",ans); 44 } 45 return 0; 46 }