《挑战》2.1 POJ POJ 1979 Red and Black (简单的DFS)
B - Red and Black
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) 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
题目大意:
就是说,对于一个h*w的网格,每次从‘@’开始,求最多能走多少个‘.’,每次只能够向四个方向进行扩展。
解题思路:
直接dfs,然后用book数组随时记录目前这个点有没有走过,如果这个点是'.'并且没有走过的话,那么我们就res++,由于是多组数据,每次输出结束后,记着memset,
复杂度:
O(4*w*h)
代码:
1 # include<cstdio> 2 # include<iostream> 3 # include<cstring> 4 5 using namespace std; 6 7 # define MAX 23 8 9 char grid[MAX][MAX]; 10 int book[MAX][MAX]; 11 int nxt[4][2] = {{1,0},{0,-1},{-1,0},{0,1} }; 12 int res; 13 int w,h; 14 15 int can_move ( int x,int y ) 16 { 17 if ( x>=0&&x<h&&y>=0&&y<w ) 18 { 19 if ( grid[x][y]=='.'&&book[x][y]==0 ) 20 return 1; 21 } 22 return 0; 23 } 24 25 void dfs ( int x,int y ) 26 { 27 book[x][y] = 1; 28 for ( int i = 0;i < 4;i++ ) 29 { 30 int n_x = x+nxt[i][0], n_y = y+nxt[i][1]; 31 if ( can_move ( n_x,n_y ) ) 32 { 33 res++; 34 book[n_x][n_y] = 1; 35 dfs(n_x,n_y); 36 } 37 } 38 } 39 40 int main(void) 41 { 42 while ( scanf("%d%d",&w,&h)!=EOF ) 43 { 44 res = 1; 45 if ( w==0&&h==0 ) 46 break; 47 for ( int i = 0;i < h;i++ ) 48 scanf("%s",grid[i]); 49 for ( int i = 0;i < h;i++ ) 50 { 51 for ( int j = 0;j < w;j++ ) 52 { 53 if ( grid[i][j]=='@' ) 54 { 55 book[i][j] = 1; 56 dfs(i,j); 57 } 58 } 59 } 60 61 printf("%d\n",res); 62 memset(book,0,sizeof(book)); 63 } 64 65 66 67 return 0; 68 }