Red and Black HDU - 1312
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.
InputThe 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)
OutputFor 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
45 59 6 13
思路:BFS
AC Code:
#include<iostream> #include<cstdio> #include<queue> #include<utility> #include<cstring> using namespace std; typedef pair<int,int> P; char maps[25][25]; int vis[25][25]; int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1}; int W,H,cnt; void bfs(P p){ queue<P> q;q.push(p); while(!q.empty() ){ p=q.front() ;q.pop() ; int x=p.first,y=p.second; for(int i=0;i<4;i++) { int nx=x+dx[i],ny=y+dy[i]; if(nx>=0&&nx<H&&ny>=0&&ny<W&&maps[nx][ny]!='#'&&vis[nx][ny]==0){ cnt++; vis[nx][ny]=1; q.push(P(nx,ny)); } } } } int main(){ while(~scanf("%d%d",&W,&H)){ if(W==0&&H==0) break; int x,y; getchar(); memset(vis,0,sizeof(vis)); for(int i=0;i<H;i++){ for(int j=0;j<W;j++){ scanf("%c",&maps[i][j]); if(maps[i][j]=='@') { x=i; y=j; } } getchar(); } cnt=1;vis[x][y]=1; bfs(P(x,y)); printf("%d\n",cnt); } }
人生不如意的时候,是上帝给的长假,这个时候应该好好享受假期。
突然有一天假期结束,时来运转,人生才是真正开始了。
突然有一天假期结束,时来运转,人生才是真正开始了。