hdu1312

Red and Black

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

Total Submission(s): 13457 Accepted Submission(s): 8334



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

本体就是一个简单的dfs深搜 变搜索边计数

dfs 深度优先搜索

//hdu1312
#include <iostream>

using namespace std;
char a[50][50];
int m,n;
int sum;
void dfs(int i,int j){
    a[i][j]='#';//把搜素到的点变成#
    sum++;//sum用于计数每搜到一个点就记一下数
    if(i-1>=0&&a[i-1][j]=='.')dfs(i-1,j);
    if(i+1<m&&a[i+1][j]=='.')dfs(i+1,j);
    if(j-1>=0&&a[i][j-1]=='.')dfs(i,j-1);
    if(j+1>=0&&a[i][j+1]=='.')dfs(i,j+1);//四个方向搜素
}
int main()
{
    while(cin>>n>>m&&(m+n)){
        for(int i=0; i<m; i++)
        cin>>a[i];
        sum=0;
        for(int i=0; i<m; i++)
        for(int j=0; j<n; j++){
            if(a[i][j]=='@')
                dfs(i,j);
        }
        cout<<sum<<endl;
    }
    return 0;
}