HDU 1010：Tempter of the Bone（DFS+奇偶剪枝+回溯）

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 146511    Accepted Submission(s): 39059

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

题意

给出一个N*M的矩阵和一个时间T，问能不能在时间恰好为T的时候从S走到D。X不能走'.'可以走，每次走过之后'.'就会消失。

AC代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#define ll long long
#define ms(a) memset(a,0,sizeof(a))
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
const double E=exp(1);
const int maxn=1e2+10;
using namespace std;
char ch[maxn][maxn];
int vis[maxn][maxn];
int ans;
int xx,yy;
int T,n,m;
int flag;
int dir[4][2]={1,0,-1,0,0,1,0,-1};
void dfs(int x,int y,int t)
{
	vis[x][y]=1;
	//如果正好在T时刻走到D
	if(t==T&&ch[x][y]=='D')
	{
		flag++;
		return ;
	}
	//奇偶剪枝，如果相差的时间和相差的曼哈顿距离的奇偶性不同，则一定无法到达
	//或者相差的时间小于曼达顿距离也不行
	//PS:不剪枝会超时
	int res=T-t-abs(xx-x)-abs(yy-y);
	if(res<0||res%2)
		return ;
	for(int i=0;i<4;i++)
	{
		int dx=x+dir[i][0];
		int dy=y+dir[i][1];
		if(dx>=0&&dx<n&&dy>=0&&dy<m&&ch[dx][dy]!='X'&&vis[dx][dy]==0)
		{
			dfs(dx,dy,t+1);
			//如果能够走到D，就可以结束了，不需要回溯
			if(flag)
				return ;
			// 回溯
			vis[dx][dy]=0;
		}
	}
}
int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false);
	while(cin>>n>>m>>T&&T&&n&&m)
	{
		ms(vis);
		flag=0;
		ans=0;
		int x,y;
		for(int i=0;i<n;i++)
			cin>>ch[i];
		// 记录开始和结束的位置
		for(int i=0;i<n;i++)
			for(int j=0;j<m;j++)
			{
				if(ch[i][j]=='S')
				{x=i;y=j;}
				if(ch[i][j]=='D')
				{xx=i;yy=j;}
			}

		dfs(x,y,0);
		if(flag)
			cout<<"YES"<<endl;
		else
			cout<<"NO"<<endl;
	}
	return 0;
}