搜索算法11之1016

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1 题目编号：1016

2 题目内容：

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<br>....#.<br>.....#<br>......<br>......<br>......<br>......<br>......<br>#@...#<br>.#..#.<br>11 9<br>.#.........<br>.#.#######.<br>.#.#.....#.<br>.#.#.###.#.<br>.#.#..@#.#.<br>.#.#####.#.<br>.#.......#.<br>.#########.<br>...........<br>11 6<br>..#..#..#..<br>..#..#..#..<br>..#..#..###<br>..#..#..#@.<br>..#..#..#..<br>..#..#..#..<br>7 7<br>..#.#..<br>..#.#..<br>###.###<br>...@...<br>###.###<br>..#.#..<br>..#.#..<br>0 0<br>

Sample Output

45<br>59<br>6<br>13<br>

Source

Asia 2004, Ehime (Japan), Japan Domestic

3 解题思路形成过程：一个人从@点出发，求他所能到达的'.'的数目，'#'不可走，@本身算1个点。比较经典的一个深搜题，首先遇到障碍或边界时退出，此时没
有通路，因此返回的道路条数是0；若可以继续走时，把走过的地方设置成障碍，防止再走一遍，这条路可以走完，然后从四个方向再继续走。

4 代码:

#include<iostream>
using namespace std;
char mapp[65][65];
int m, n, k = 0, c = 0, d = 0;
void dfs(int x, int y)
{
if (mapp[x][y] == '#' || x<0 || y<0 || x >= m || y >= n)
{
return;
}
mapp[x][y] = '#';
dfs(x + 1, y);
dfs(x - 1, y);
dfs(x, y + 1);
dfs(x, y - 1);
}
int main()
{
int i, j, k;
while (cin >> n >> m)
{
c = 0;
d = 0;
if (m == 0 && n == 0)
break;
for (i = 0; i<m; i++)
{
for (j = 0; j<n; j++)
{
cin >> mapp[i][j];
}
}
for (i = 0; i<m; i++)
for (j = 0; j<n; j++)
{
if (mapp[i][j] == '#')
c++;
}
k = 0;
for (i = 0; i<m; i++)
{
for (j = 0; j<n; j++)
{
if (mapp[i][j] == '@')
{
mapp[i][j] = '.';
dfs(i, j);
}
}
}
for (i = 0; i<m; i++)
for (j = 0; j<n; j++)
{
if (mapp[i][j] == '#')
d++;
}
cout << d - c << endl;
}
return 0;
}

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