POJ 1979 Red and Black(dfs)

Red and Black

Time Limit: 1000MS

Memory Limit: 30000K

Total Submissions: 27110

Accepted: 14737

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

4559613

题意：在一块N*M的矩形房间里，有黑色'.'和红色'#'两种地砖，现在XX站在一块黑色地砖'@'上，他只能向前后左右四个方向移动，且不能移动到红色瓷砖上。问XX能最多走多少块地砖？

dfs简单题，代码如下：

#include<cstdio>#include<cstring>int n,m,ans;char map[25][25];int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};void dfs(int x,int y){map[x][y]='#';ans++;for(int i=0;i<4;++i){int nx=x+dir[i][0];int ny=y+dir[i][1];if(0<=nx&&nx<n&&0<=ny&&ny<m&&map[nx][ny]=='.')dfs(nx,ny);}return ;} int main(){int i,j,x,y;while(scanf("%d%d",&m,&n)&&n||m){for(i=0;i<n;++i){scanf("%s",map[i]);for(j=0;j<m;++j){if(map[i][j]=='@'){x=i;y=j;}}}ans=0;dfs(x,y);printf("%d\n",ans);}}