POJ1979（Red and Black）

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 18460		Accepted: 9794

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)
The end of the input is indicated by a line consisting of two zeros.

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
没什么说的，简单的递归搜索。。。

//POJ 1979	Accepted	184K	16MS	C++	1052B	2013-04-11 19:54:05
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
using namespace std;
const int maxn = 25;

string map1[maxn];//学习string的用法。
bool vis[maxn][maxn];
int n, m;
int ans = 0;

void search1(int row, int col) {
    if(row<0 || row >= m||col<0 || col>=n ||map1[row][col]=='#'||vis[row][col]) return;
    ++ans;
    vis[row][col] = true;
    search1(row+1, col);
    search1(row-1, col);
    search1(row, col+1);
    search1(row, col-1);
}


int main()
{
    int row, col;
    while(scanf("%d%d", &n, &m) != EOF) {//先输入列再输入行。
        if(m == 0&&n==0) break;
        memset(vis, false,sizeof(vis));
        ans = 0;
        for(int i = 0; i < m; i++) {
            cin >> map1[i];             //#include <string>
            for(int j = 0; j < n; j++) {
                if(map1[i][j]=='@') {
                    row = i;
                    col = j;
                }
            }
        }
        search1(row, col);
        printf("%d\n", ans);
    }
    return 0;
}