POJ 1979 Red and Black
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Red and Black
Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 33405 Accepted: 18176
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.
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 Output
4559613
Source
Japan 2004 Domestic
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 using namespace std; 5 char map[25][25]; 6 bool vis[25][25]; 7 int n,m,sx,sy,sum; 8 int dx[]={0,0,1,-1}; 9 int dy[]={1,-1,0,0};10 void Dfs(int x,int y){11 sum++;vis[x][y]=true;12 for(int i=0;i<4;i++){13 int nx=x+dx[i],ny=y+dy[i];14 if(nx<=n&&nx>=1&&ny<=m&&ny>=1&&15 map[nx][ny]=='.'&&!vis[nx][ny])16 Dfs(nx,ny);17 }18 }19 void init(){20 memset(vis,false,sizeof(vis));21 memset(map,'0',sizeof(map));22 sum=0;23 }24 int main()25 {26 while(1){27 scanf("%d%d",&m,&n);28 if(n==0&&m==0) break;29 init();30 for(int i=1;i<=n;i++)31 for(int j=1;j<=m;j++){32 cin>>map[i][j];33 if(map[i][j]=='@') sx=i,sy=j;34 }35 Dfs(sx,sy);36 printf("%d\n",sum);37 }38 return 0;39 }
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