【hdu1010】Tempter of the Bone——dfs

题目：

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 98266    Accepted Submission(s): 26642

Problem 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.

 

Input

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.

 

Sample Input

  

   4 4 5
S.X.
..X.
..XD
....
3 4 5
S.X.
..X.
...D
0 0 0
  

 

Sample Output

  

   NO
YES
  

 

Author

ZHANG, Zheng

 

Source

ZJCPC2004

 

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描述：走迷宫，在 规定步数内走到终点才算是通过。

题解：我采用的方法是步数作为参数直接传下去，不开标记数组，当dfs走过一点可以继续往下走时就把该点标记为X；要注意的是首先终点也会标记为.（回溯时），因此不能通过某点的map值来判断是否为终点，而是要通过比较之前记录的终点坐标，其次要么在一开始的时候把起点标记为X，要么在查询是合法范围里不包含起点；最重要的一点——剪枝！即可以dfs的条件，因为最短路径就是起终点坐标差绝对值之和，如果这个长度与T相差为偶数，那么如果提前到达就有可能通过多走两步来“拖延时间”，如向右走一步变为上右下，但是奇数的话是回不来的，所以这样就不用dfs，直接输出NO。

代码：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
using namespace std;
char map[10][10];
int dx[] = { 0,0,1,-1 };
int dy[] = { 1,-1,0,0 };
int N, M, T,endx,endy;
void init()
{
	memset(map, 0, sizeof(map));
}

bool dfs(int depth, int x, int y)
{
	if (depth == T || (x == endx && y == endy))
		return depth == T && (x == endx && y == endy);

	for (int i = 0; i < 4; i++)
	{
		int curx = x + dx[i];
		int cury = y + dy[i];
		if (curx >= 0 && curx < N && cury >= 0 && cury < M && (map[curx][cury] == 'D' || map[curx][cury] == '.'))
		{
			if (map[curx][cury] == '.')
				map[curx][cury] = 'X';
			if (dfs(depth + 1, curx, cury))
				return true;
			else
				map[curx][cury] = '.';
		}
	}
	return false;
}

int main()
{
	//freopen("input.txt", "r", stdin);
	while (cin>>N>>M>>T)
	{
		if (N == 0 && M == 0 && T == 0)
			break;
		init();
		int startx, starty;
		for (int i = 0; i < N; i++)
		{
			for (int j = 0; j < M; j++)
			{
				cin >> map[i][j];
				
				if (map[i][j] == 'S')
				{
					startx = i;
					starty = j;
				}
				else if (map[i][j] == 'D')
				{
					endx = i;
					endy = j;
				}
			}
		}

		int temp = abs(abs(endx - startx) + abs(endy - starty)-T);
		bool ans = false;
		if (temp % 2 == 0)
			ans = dfs(0, startx, starty);
		
		if (ans)
			cout << "YES" << endl;
		else
			cout << "NO" << endl;
	}

	return 0;
}