POJ 1979 Red and Black（紅與黑）

POJ 1979 Red and Black（紅與黑）

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.

'.' - 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

分析

題目中有如下要求：

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

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。

和走的方向在於下方代碼中定義的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

#include       
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;
    }
}

號外

求投票或轉發支持呀……希望我不要死得太慘了……

請點擊這裡：投票

投票從10號開始一直持續到20號，拜托各位了！

———————————————————當然你也可以直接點擊圖片啦