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)
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
题意:找出@能够到达的所有点的个数
思路:dfs和递归都能做
dfs代码:
1 #include <iostream> 2 #include<cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <queue> 6 using namespace std; 7 char s[25][25]; 8 int a[4][2]={{1,0},{-1,0},{0,1},{0,-1}}; 9 int ans,n,m; 10 int dfs(int x,int y) 11 { 12 for(int i=0;i<4;i++) 13 { 14 int tx=x+a[i][0]; 15 int ty=y+a[i][1]; 16 if(tx>=m||tx<0||ty<0||ty>=n) 17 continue; 18 if(s[tx][ty]=='.') 19 { 20 s[tx][ty]='#'; 21 dfs(tx,ty); 22 ans++; 23 } 24 } 25 return 0; 26 } 27 int main() 28 { 29 while(~scanf("%d%d",&n,&m)) 30 { 31 if(n==0||m==0) 32 break; 33 int flag=1; 34 int tx,ty; 35 for(int i=0;i<m;i++) 36 { 37 scanf("%s",s[i]); 38 if(flag) 39 { 40 for(int j=0;j<n;j++) 41 { 42 if(s[i][j]=='@') 43 { 44 tx=i,ty=j; 45 flag=0; 46 break; 47 } 48 } 49 } 50 } 51 ans= 1; 52 dfs(tx,ty); 53 printf("%d ",ans); 54 } 55 return 0; 56 }
递归代码:
1 #include <iostream> 2 #include <algorithm> 3 #include <cstring> 4 #include <stdio.h> 5 int n,m; 6 char data[30][30]; 7 using namespace std; 8 int f(int i,int j) 9 { 10 if(i<0||j>n-1||i>m-1||j<0) 11 return 0; 12 if(data[i][j]=='#') 13 return 0; 14 data[i][j]='#'; 15 return 1+f(i-1,j)+f(i+1,j)+f(i,j-1)+f(i,j+1); 16 } 17 int main() 18 { 19 int x,y; 20 while(~scanf("%d%d",&n,&m)&&(n!=0||m!=0)) 21 { 22 getchar(); 23 for(int i=0;i<m;i++) 24 scanf("%s",data[i]); 25 for(int i=0;i<m;i++) 26 for(int j=0;j<n;j++) 27 if(data[i][j]=='@') 28 { 29 x=i;y=j; 30 } 31 cout<<f(x,y)<<endl; 32 } 33 return 0; 34 }