hdu 1312 Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 9949    Accepted Submission(s): 6193

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
跟hdu 1241思路差不多
#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespace std;int d[4][2]={{0,-1},{0,1},{-1,0},{1,0}};char map[25][25];int n,m,sum;void dfs(int x,int y){     if(map[x][y]=='#') return ;     sum++;     map[x][y]='#';     for(int i=0;i<4;i++)     {             int xx=x+d[i][0],yy=y+d[i][1];             if(xx>=0 && xx<m && yy>=0 && yy<n)                dfs(xx,yy);     }}int main(){    int x,y;    while(cin>>n>>m)    {         if(n==0 || m==0)  break;         getchar();         for(int i=0;i<m;i++)         {            gets(map[i]);            for(int j=0;j<n;j++)                if(map[i][j]=='@')                      x=i,y=j;         }         sum=0;                  dfs(x,y);         printf("%d\n",sum);    }    return 0;}