Red and Black（dfs水题）

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Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21071    Accepted Submission(s): 12843

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

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

题意：一个人从起点只能走黑色地板，只能上下左右移动，计算最多可以走多少块。

题解：dfs水题，直接贴代码。

#include<iostream>#include<cstring>#include<cstdio>using namespace std;int n,m,vis[30][30],ans;int fx[4]={0,0,1,-1},fy[4]={1,-1,0,0};char tile[30][30];void dfs(int x,int y){ans++;//计算步数vis[x][y]=1;for(int i=0;i<4;i++){int nx=x+fx[i];int ny=y+fy[i];if(nx>=0&&nx<n&&ny>=0&&ny<m&&!vis[nx][ny]&&tile[nx][ny]=='.')dfs(nx,ny);}}int main(){while(cin>>m>>n&&(m&&n)){ans=0;getchar();memset(vis,0,sizeof(vis));for(int i=0;i<n;i++)cin>>tile[i];for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(tile[i][j]=='@'&&!vis[i][j]){dfs(i,j);break;}}}cout<<ans<<endl;}return 0;}