Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 72102    Accepted Submission(s): 19839

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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普通搜索不行，需要奇偶剪枝，奇偶剪枝的意思是如果当前点到目标点的最少步数的奇偶性和剩余时间的奇偶性不同就直接return，因为奇数步需要奇数时间，偶数步需要偶数时间

#include<map>
#include<set>
#include<list>
#include<stack>
#include<queue>
#include<vector>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

int dir[4][2] = { {1, 0}, {-1, 0}, {0, 1}, {0, -1} };
bool vis[10][10];
char mat[10][10];
int n, m, t;
int sx, sy, ex, ey;
bool flag;

bool is_legal(int x, int y)
{
	if(x < 0 || x >= n || y < 0 || y >= m || mat[x][y] == 'X')
		return false;
	return true;
}

void dfs(int x, int y, int step)
{
	if(flag)
		return;
	if(t - step < (abs(x - ex) + abs(y - ey)))
		return;
	if((abs(x - ex) + abs(y - ey)) % 2 != (t - step) % 2)
		return ;
	if(mat[x][y] == 'D')
	{
		if(step == t)
			flag = true;
		return;
	}
	for(int i = 0; i < 4; i++)
	{
		int xx = x + dir[i][0];
		int yy = y + dir[i][1];
		if( !is_legal(xx, yy) )
			continue;
		if(vis[xx][yy])
			continue;
		vis[xx][yy] = 1;
		dfs(xx, yy, step + 1);
		vis[xx][yy] = 0;
	}
}

int main()
{
	while(~scanf("%d%d%d", &n, &m, &t))
	{
		if( !t && !n && !m)
			break;
		int wall = 0;
		for(int i = 0; i < n; i++)
		{
			scanf("%s", mat[i]);
			for(int j = 0; j < m; j++)
			{
				if(mat[i][j] == 'S')
				{
					sx = i;
					sy = j;
				}
				if(mat[i][j] == 'D')
				{
					ex = i;
					ey = j;
				}
				if(mat[i][j] == 'X')
				{
					wall ++;
				}
			}
		}
		if(n * m - wall < t)
		{
			printf("NO\n");
			continue;
		}
		int s = sx * n + sy;
		int e = ex * n + ey;
		//printf("%d %d %d %d\n", sx, sy, ex, ey);
		memset( vis, 0, sizeof(vis) );
		vis[sx][sy] = 1;
		flag = false;
		dfs(sx, sy, 0);
		if(flag)
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}