原文

    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.

<code class="hljs livecodeserver">'.' - a black tile '#' - a red tile '@' - a man on a black tile(appears exactly once in a data set)</code>

    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

<code class="hljs ruleslanguage">6 9....#......#..............................#@...#.#..#.11 9.#..........#.#######..#.#.....#..#.#.###.#..#.#..@#.#..#.#####.#..#.......#..#########............11 6..#..#..#....#..#..#....#..#..###..#..#..#@...#..#..#....#..#..#..7 7..#.#....#.#..###.###...@...###.###..#.#....#.#..0 0</code>

    Sample Output

<code class="hljs ">4559613</code>

分析

    題目中有如下要求：

<code class="hljs ">只能走周圍的4個相鄰點只能走黑色點，不能走紅色點一次只能走一點</code>

    需要計算的是：能走到的黑色點的和

    因為”.“表示黑色點，所以在下面的dfs函數中需要判斷當前點為黑色點才可以進行下一步搜索，

POJ 1979 Red and Black（紅與黑）

電腦資料

    和走的方向在于下方代碼中定義的direc二維數組。

    而其具體的方向等信息，我全都列在下圖了。

代碼

<code class="hljs cpp">#include<iostream>using namespace std;// 題目中給出的最大寬度和高度#define MAX_W 20#define MAX_H 20// 待輸入的寬度和高度以及已走的步數int W, H;     int step = 0;// 待寫入的二維數組char room[MAX_W][MAX_H];// 順時針的可走方向const int direc[4][2] = {    {0, -1},    {1,0},    {0, 1},    {-1 ,0},};int dfs(const int& row, const int& col) {    // 走過的點    room[row][col] = '#';    // 計算步數    ++step;    for (int d = 0; d < 4; ++d) {        int curRow = row + direc[d][1];        int curCol = col + direc[d][0];        if (curRow >= 0 && curRow < H && curCol >= 0 && curCol < W && room[curRow][curCol] == '.') {            dfs(curRow, curCol);        }    }    return step;}int main(){    bool found;    while (cin >> W >> H, W > 0) {        step = 0;        int col, row;        // 輸入        for (row = 0; row < H; ++row) {            for (col = 0; col < W; ++col) {                cin >> room[row][col];            }        }        found = false;        // 找到起始點        for (row = 0; row < H; ++row) {            for (col = 0; col < W; ++col) {                if (room[row][col] == '@') {                    found = true;                    break;                }            }            if (found) {                break;            }                           }        // 開始搜索        cout << dfs(row, col) << endl;    }}</iostream></code>

號外

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