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poj 1979 Red and Black
2015-07-20 17:59:43 来源: 作者: 【 】 浏览:1
Tags:poj 1979 Red and Black
Red and Black
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 22300 Accepted: 12041

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.

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.

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

Japan 2004 Domestic

题意:
首先这是一个图,' . '代表黑砖,‘#’代表红砖,‘@’表示主人公呆的那块黑砖。红砖不能走,只能沿着黑砖走,每次走一步,问能走到的黑砖的最大数目!

金典DFS;
注意边界,递归结束条件!

上代码:
#include
  
   
#include
   
     #include
    
      #include
     
       using namespace std; #define maxn 25 #define MM 10000000 char s[maxn][maxn]; int w[maxn][maxn]; //标记走过的点 int sx,sy; int conut; //计数 int n,m; int xx[]={0,0,1,-1}; int yy[]={1,-1,0,0}; int dfs(int x,int y) { if(s[x][y]=='#' || w[x][y]) return 0; conut++; w[x][y]=1; for(int i=0;i<4;i++) { int xi=x+xx[i]; int yi=y+yy[i]; if(xi>=0 && xi
      
       =0 && yi
       
        

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