HDU 1010 Tempter of the Bone

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本文介绍了一款迷宫逃脱游戏的设计与实现。游戏中的狗狗需要在一个大小为N*M的矩形迷宫中找到出口。迷宫包含墙壁、起点、终点及空地等元素。狗狗必须在门开启的瞬间到达门的位置才能成功逃脱。文章提供了使用深度优先搜索(DFS)算法解决此问题的详细步骤及代码实现。

Description

The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze began to shake, and the doggie could feel the ground sinking. He realized that the bone was a trap, and he tried desperately to get out of this maze.

The maze was a rectangle with sizes N by M. There was a door in the maze. At the beginning, the door was closed and it would open at the T-th second for a short period of time (less than 1 second). Therefore the doggie had to arrive at the door on exactly the T-th second. In every second, he could move one block to one of the upper, lower, left and right neighboring blocks. Once he entered a block, the ground of this block would start to sink and disappear in the next second. He could not stay at one block for more than one second, nor could he move into a visited block. Can the poor doggie survive? Please help him.

I/O

The input consists of multiple test cases. The first line of each test case contains three integers N, M, and T (1 < N, M < 7; 0 < T < 50), which denote the sizes of the maze and the time at which the door will open, respectively. The next N lines give the maze layout, with each line containing M characters. A character is one of the following:

‘X’: a block of wall, which the doggie cannot enter;
‘S’: the start point of the doggie;
‘D’: the Door; or
‘.’: an empty block.

The input is terminated with three 0’s. This test case is not to be processed.

Output
For each test case, print in one line “YES” if the doggie can survive, or “NO” otherwise.

Algorithm

DFS

Hint

DFS模板题，有几点注意。

getchar()消去输入中的换行符
开一个大数组，从maze[1][1]开始读取，并初始化为X，可避免搜索时判断边界

剪枝。不剪枝是会TLE的，具体剪枝：

int temp  = 0; 

temp = len- abs(dx - x) - abs(dy - y); 

if(temp < 0 || temp & 1) 

    return;

if((abs(dx-sx)+abs(dy-sy))%2 != (t)%2) 

        printf(“NO\n”); 

均通过判断步长奇偶性来实现。

Code

#include<cstdio>
#include<cmath>
#include<cstring>

int dir_x[4] = {1,-1,0,0};
int dir_y[4] = {0,0,1,-1};
int n, m, t, sx, sy, dx, dy;
char maze[9][9];
bool flag;

void dfs(int x, int y,int len)
{
    int temp  = 0;
    temp = len- abs(dx - x) - abs(dy - y);
    if(temp < 0 || temp & 1)
        return;
    for(int i = 0; i<4 && flag == 0 && len>0; i++)
    {
        if(maze[x+dir_x[i]][y+dir_y[i]] == '.')
        {
            maze[x+dir_x[i]][y+dir_y[i]] = 'X';
            dfs(x+dir_x[i], y+dir_y[i], len-1);
            maze[x+dir_x[i]][y+dir_y[i]] = '.';
        }
        if(x+dir_x[i] == dx && y+dir_y[i] == dy && len == 1)
        {
            flag = 1;
            return;
        }
    }
    return;
}

int main()
{
    while(scanf("%d%d%d",&n, &m, &t) == 3)
    {
        getchar();
        memset(maze, 'X', sizeof(maze));
        if(n+m+t == 0)
            break;
        for(int i = 1; i<=n; i++)
        {
            for(int j = 1; j<=m; j++)
            {
                scanf("%c",&maze[i][j]);
                if(maze[i][j] == 'S')
                {
                    sx = i;
                    sy = j;
                }
                if(maze[i][j] == 'D')
                {
                    dx = i;
                    dy = j;
                }
            }
            getchar();
        }
        flag = 0;
        if((abs(dx-sx)+abs(dy-sy))%2 != (t)%2)
            printf("NO\n");
        else
        {
            maze[sx][sy] = 'X';
            dfs(sx, sy, t);
            if(flag)
                printf("YES\n");
            else
                printf("NO\n");
        }
    }
    return 0;
}