題目描述

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.

輸入

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.

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

解題思路：
深搜的方法解決，題目意思就是從@開始找.并與@連通，碰到#等于碰到了墻，題目很簡(jiǎn)單，@可以向四個(gè)方向上、下、左、右走，所以 用四個(gè)坐標(biāo)標(biāo)記出來(lái)，然后，再一一遍歷，遞歸調(diào)用尋找，用一個(gè)30*30的數(shù)組標(biāo)識(shí)此點(diǎn)有沒(méi)有走過(guò)，避免走重復(fù)

程序代碼： #include <cstdio> #include <cstring> using namespace std; int n,m,cot; char map[30][30]; int to[4][2] = {{1,0},{0,1},{-1,0},{0,-1}};void dfs(int i,int j) {cot++;map[i][j] = '#';for(int k = 0; k<4; k++){int x = i+to[k][0];int y = j+to[k][1];if(x<n && y<m && x>=0 && y>=0 && map[x][y] == '.')dfs(x,y);}return; }int main() {int i,j,fi,fj;while(~scanf("%d%d%*c",&m,&n)&&m&&n){for(i = 0; i<n; i++){for(j = 0; j<m; j++){scanf("%c",&map[i][j]);if(map[i][j] == '@'){fi = i;fj = j;}}getchar();}cot= 0;dfs(fi,fj);printf("%d\n",cot);}return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/www-cnxcy-com/p/4678523.html

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