【JAVA】HDU 1312 Red and Black（DFS水题）

先上原题：

                                Red and Black                 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)                 Total Submission(s): 21177    Accepted Submission(s): 12911

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

一道很简单的DFS水题，不必考虑剪枝可以直接过，题意大致是：
在给定的有起点的房间里面，一个可以通过’.’而不能通过’#’的人能到达多少方格，需要注意的是，起点也包含在内。

用JAVA暴力深搜一遍AC了，代码如下：

import java.util.Scanner;public class Main{    static int w = 0;//每个测试房间的宽度    static int h = 0;//每个测试房间的高度    static char room[][] = new char[21][21];//房间    static int sum = 1;//计算可到达的方格    public static void main(String args[]){        Scanner sc = new Scanner(System.in);        while(sc.hasNext()){            w = sc.nextInt();            h = sc.nextInt();            if(w == 0 && h == 0)break;            String a = "";            for(int i = 0;i < h;i++){                a = sc.next();                room[i] = a.toCharArray();            }            //遍历房间寻找起点            int x = 0,y = 0;            for(int i = 0;i < h;i++){                for(int j = 0;j < w;j++){                    if(room[i][j] == '@'){                        x = j;                        y = i;                    }                }            }            dfs(x,y);            System.out.println(sum);            sum = 1;        }    }    static void dfs(int x, int y){        if( y >= 0 && x >= 0 && y < h && x < w){        if(room[y][x] == '#')return;//不可前进时停止            if(room[y][x] == '.')                ++sum;//计数        }        else return;//到达房间边缘时停止        room[y][x] = '#';//将走过的地板变为不可行走        //继续前进        dfs(x-1,y);        dfs(x+1,y);        dfs(x,y-1);        dfs(x,y+1);    }}