POJ--1979 Red and Black(DFS)

记录
22:42 2023-2-3

http://poj.org/problem?id=1979

reference:《挑战程序设计竞赛(第2版)》第二章练习题索引 p135

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

or 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

一个dfs，注意遍历的时候要将走过的地方修改，不然可能会重复走。

#include<cstdio>
#define MAX_W 20
#define MAX_H 20
char tiles[MAX_H + 1][MAX_W + 1];

int W, H;
int start_x, start_y;
int new_x, new_y;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int res = 1;

void dfs(int x, int y) {
    tiles[x][y] = '#';
    for(int i = 0; i < 4; i++){
        new_x = x + dx[i];
        new_y = y + dy[i];
        if(0 <= new_x && new_x < H && 0 <= new_y && new_y < W && tiles[new_x][new_y] == '.') { 
            res += 1;
            dfs(new_x, new_y);
        }
    } 
}

void solve() {
    res = 1;
    dfs(start_x, start_y);
    printf("%d\n", res);
}

int main() {
    while (~scanf("%d %d\n", &W, &H)) {
        if(W == 0 && H == 0) {
            return 0;
        }
        for(int i = 0; i < H; i++) {
            for(int j = 0; j < W; j++){
                scanf("%c", &tiles[i][j]);
                if(tiles[i][j] == '@') {
                    start_x = i;
                    start_y = j;
                }
            }
            getchar();
        } 
        solve();       
    }
    return 0;
}