HDU 1312 Red and Black

Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15538 Accepted Submission(s): 9623

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

45
59
6
13

这是一题非常简单的DFS，DFS就是深度优先搜索算法;
它不是很难，有点点耐心就可以学会：
它主要运用了递归知识，这个一定要搞懂才能更好理解DFS；
用点耐心吧，加油：

#include <iostream>#include <sstream>using namespace std;char c[100][100];bool  c1[100][100];int c2[][2] = { 0,1,0,-1,1,0,-1,0 };int x, y;int s;void  NFS(int a,  int b){       c1[b][a] = true;    int dd, ddd;    for (int i = 0; i < 4; i++)    {        dd = a + c2[i][0];        ddd = b + c2[i][1];        if (dd >= 0 && dd < x&&ddd >= 0 && ddd < y&&c[ddd][dd] != '#'&&c[ddd][dd] != '@'&&!c1[ddd][dd])        {            s++;            NFS(dd, ddd);        }    }}int main( ){    while (cin >> x >> y)    {           if (x == 0 && y == 0)        {            return 0;        }        memset(c1, false, sizeof(c1));        int x1, y1;        for (int i = 0; i < y; i++)        {            for (int j = 0; j < x; j++)            {                cin >> c[i][j];                if(c[i][j]=='@')                {                    x1 = j;                    y1 = i;                }            }        }        s = 1;        NFS(x1, y1);        cout << s << endl;    }    return 0;}