Red and Black 【hdu-1312】【dfs】

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21103    Accepted Submission(s): 12862

Problem 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) 

 

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

 

 题解：dfs水题。

代码如下：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int H,W;int st_x,st_y;int ans;char mapp[25][25];int dx[4]={0,0,-1,1},dy[4]={1,-1,0,0};void dfs(int x,int y){ans++;mapp[x][y]='#';for(int i=0;i<4;i++){int nx=x+dx[i];int ny=y+dy[i];if(0<=nx&&nx<H&&0<=ny&&ny<W&&mapp[nx][ny]=='.')dfs(nx,ny);}}int main(){while(scanf("%d%d",&W,&H)){if(W==0&&H==0) break;ans=0;for(int i=0;i<H;i++){scanf("%s",mapp[i]);}int q=0;for(int i=0;i<H;i++){for(int j=0;j<W;j++){if(mapp[i][j]=='@'){st_x=i,st_y=j;q=1;break;}}if(q==1) break;}dfs(st_x,st_y);printf("%d\n",ans);}return 0;}