【FZUOJ 1046】Tempter of the Bone(DFS加奇偶剪枝)

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于 2018-05-16 22:53:14 发布

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分类专栏：搜索--DFS

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本文介绍了一个迷宫逃脱问题，通过深度优先搜索（DFS）结合剪枝策略解决狗仔如何在限定时间内找到出口的问题。该算法考虑了迷宫的大小、门开启的时间及狗仔的移动限制。

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Accept: 623 Submit: 1734
Time Limit: 1000 mSec Memory Limit : 32768 KB

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 5S.X...X...XD....3 4 5S.X...X....D0 0 0

Sample Output

NOYES

Source

ZOJ 2110

我们如果想从(x,y)走到终点(dx,dy)，最短的距离应当为abs(dx-x)+abs(dy-dy),假设这个距离为D1，如果存在D2能走到终点，那么D2的奇偶性一定和D1相同，因为起点相同，终点相同如果在横轴上面加1,那么在纵轴上面也应该加1，所以我们在深搜之前可以先进行剪枝一下。这样时间复杂度会降低很多。

#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
using namespace std;
char maze[100][100];
int n,m,T;
int flag;
int sx,sy,dx,dy;
int vis[100][100];
int dir[4][2] = {{1,0},{0,1},{-1,0},{0,-1}};
void dfs(int x,int y,int t)
{
	int nx,ny;
	if(x==dx&&y==dy&&t==T)
	{
		flag=1;
		return ;
	}
	if(x<0&&x>=n||y<0||y>=m) return ;
	int temp1=abs(dx-x)+abs(dy-y);
	int temp2=abs(T-t);
	int temp=abs(temp1-temp2);//剪枝 
	if(temp&1) return ;
	for(int i=0;i<4;i++)
	{
		nx=x+dir[i][0];
		ny=y+dir[i][1];
		if(maze[nx][ny]!='X'&&!vis[nx][ny]&&nx>=0&&nx<n&&ny>=0&&ny<m)
		{
			vis[nx][ny]=1;
			dfs(nx,ny,t+1);
			if(flag) return ;
			vis[nx][ny]=0;//回溯 
		}
	}
	
}
int main()
{
	while(cin>>n>>m>>T)
	{
		flag=0;
		memset(vis,0,sizeof(vis));
		if(n+m+T==0) break;
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<m;j++)
			{
				cin>>maze[i][j];
			}
		}
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<m;j++)
			{
				if(maze[i][j]=='D')
				{
					dx=i;
					dy=j;
				}
				else if(maze[i][j]=='S')
				{
					sx=i;
					sy=j;
					vis[i][j]=1;
				}
			}
		}
		dfs(sx,sy,0);
		if(flag) cout<<"YES"<<endl;
		else cout<<"NO"<<endl;
	}
	return 0;
}