【DFS】poj 1979 Red and Black

Red and Black

Time Limit: 1000MS
Memory Limit: 30000KTotal Submissions: 36084
Accepted: 19554

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

4559613///AC代码
#include <iostream>#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <bitset>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <algorithm>#include <functional>#define PI acos(-1)#define eps 1e-8#define inf 0x3f3f3f3f#define debug(x) cout<<"---"<<x<<"---"<<endltypedef long long ll;using namespace std;char str[55][55];bool vis[55][55];int dix[4] = {1, -1, 0, 0};int diy[4] = {0, 0, 1, -1};int m, n, sum;bool cango(int x, int y){    return x >= 1 && x <= m && y >= 1 && y <= n;}void dfs(int x, int y){    if (vis[x][y] == true)    {        return ;    }    vis[x][y] = true;    //a[x][y]='#';    sum++;    for (int i = 0; i < 4; i++)    {        int _x = x + dix[i];        int _y = y + diy[i];        if (cango(_x, _y) && str[_x][_y] == '.')        {            dfs(_x, _y);        }    }}int main(){    while (cin >> n >> m && m + n)    {        memset(vis, false, sizeof(vis));        int x, y;        sum = 0;        for (int i = 1; i <= m; i++)        {            for (int j = 1; j <= n; j++)            {                cin >> str[i][j];                if (str[i][j] == '@')                {                    x = i;                    y = j;                }            }        }        dfs(x, y);        cout << sum << endl;    }    return 0;}