HDUOJ1312 Red and Black

最新推荐文章于 2018-12-22 18:13:12 发布

最新推荐文章于 2018-12-22 18:13:12 发布 · 110 阅读

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本文介绍了一个使用深度优先搜索（DFS）算法解决迷宫寻路问题的经典案例。在一个由黑色和红色方块组成的矩形房间中，从一个指定的黑色起点出发，寻找所有可达的黑色方块。通过递归的方式实现DFS算法，最终统计并输出可到达的所有黑色方块数量。

Red and Black

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2083Accepted Submission(s): 1356

Problem 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)

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

【解题思路】像四个方向搜索可达的.的数量，并吧访问过的标记为#

#include<iostream> #include<stdio.h> using namespace std; #define N 21 char arc[N][N]; int set[N][N], w, h; int d[4][2] = { { -1, 0 }, { 1, 0 }, { 0, 1 }, { 0, -1 } }; int k; int dfs(int x, int y) { arc[x][y] = '#'; for (int i = 0; i < 4; i++) { int px = x + d[i][0]; int py = y + d[i][1]; if (px >= 0 && py >= 0 && px < h && py < w && arc[px][py] != '#') { k++; dfs(px, py); } } return k; } int main() { int i, j; int x, y; while (cin >> w >> h && w && h) { k = 1; for (i = 0; i < h; i++) for (j = 0; j < w; j++) { cin >> arc[i][j]; if (arc[i][j] == '@') { x = i;y = j; } } cout << dfs(x, y) << endl; } return 0; }