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《POJ 1979 Red and Black（紅與黑）》(https://www.unjs.com)。
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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請點擊這里：投票
投票從10號開始一直持續到20號，拜托各位了！
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