HDU - 1312 ： Red and Black
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.
InputThe 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
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OutputFor 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
@可以通過上下左右通到的.有幾個。可以DFS也可以BFS，只需把圖遍歷即可。
1 #include<iostream> 2 #include<string.h> 3 4 using namespace std; 5 6 char a[22][22]; 7 int s; 8 9 void DFS(int x1,int y1) 10 { 11 if(a[x1][y1]==‘.‘||a[x1][y1]==‘@‘) 12 { 13 a[x1][y1]=‘#‘; 14 s++;15 DFS(x1+1,y1); 16 DFS(x1-1,y1); 17 DFS(x1,y1+1); 18 DFS(x1,y1-1); 19 return; 20 } 21 } 22 23 int main() 24 { 25 int x,y; 26 while(scanf("%d %d",&y,&x),x||y) 27 { 28 int x1,y1; 29 s=0; 30 memset(a,0,sizeof(a)); 31 for(int i=1;i<=x;i++) 32 { 33 getchar(); 34 for(int j=1;j<=y;j++) 35 { 36 scanf("%c",&a[i][j]); 37 //cout<<i<<" "<<j<<endl; 38 if(a[i][j]==‘@‘) 39 { 40 x1=i; 41 y1=j; 42 } 43 } 44 } 45 DFS(x1,y1); 46 cout<<s<<endl; 47 } 48 49 return 0; 50 }
HDU - 1312 ： Red and Black