Red and Black （简单dfs）
题目：https://vjudge.net/problem/POJ-1979
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.
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.
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 Out
45
59 6 13
题意：
从@点出发，，问能走多少个.
思路：简单dfs
代码：
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<set> #include<algorithm> #include<map> #define maxn 200005 using namespace std; int x[4]={-1,1,0,0}; int y[4]={0,0,1,-1}; char mp[1005][1005]; int vis[1005][1005]={0}; int nx,ny; int ans; int n,m; int s1,s2; void dfs(int i,int j) { if(i<0||i>n||j<0||j>m)return ; vis[i][j]=1; for(int k=0;k<4;k++){ nx=i+x[k];ny=j+y[k]; if(nx>=0&&nx<n&&ny>=0&&ny<m&&!vis[nx][ny]&&mp[nx][ny]=='.'){ans++;vis[nx][ny]=1;dfs(nx,ny);} } } int main() { while(cin>>m>>n){ if(m==0&&n==0)break; getchar(); for(int i=0;i<n;i++) for(int j=0;j<m;j++) { cin>>mp[i][j]; if(mp[i][j]=='@'){s1=i;s2=j;} } memset(vis,0,sizeof(vis)); ans=0; dfs(s1,s2); cout<<ans+1<<endl; } return 0; }
