hdu 1010 Tempter of the Bone【DFS】

最新推荐文章于 2023-10-20 20:27:55 发布

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最新推荐文章于 2023-10-20 20:27:55 发布

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本文介绍了一个迷宫逃脱问题，狗需要在限定时间内找到出口。通过深度优先搜索加剪枝策略来解决这一问题，确保狗能在门开启的瞬间到达。文章包含问题描述、样例输入输出及C语言实现代码。

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Tempter of the Bone

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

题意解析

dfs+剪枝
从起点S开始搜索，然后直到D点结束。但是注意步数要要固定为t和走过的路不能重复走这两点。因为搜索可能会超时所以就需要剪枝，奇偶剪枝，就是每一步到终点的步数奇偶性要与剩余步数奇偶性相同。

程序代码 C

#include <stdio.h>
#include <math.h>
#include <string.h>
int n,m,t,flag;
char s[10][10];
int book[10][10];
int sx,sy,dx,dy; 
int next[4][2]={{0,1},{1,0},{0,-1},{-1,0}};

void dfs(int x,int y,int step){
	//printf("%d-%d\n",step,t);
	int i,k,f,tx,ty;
	if(x==dx&&y==dy&&step==t)
		flag=1;
	if(flag==1||step>=t)
		return ;
	f=abs(x-dx)+abs(y-dy);
	if(f%2!=(t-step)%2)
		return ;
	for(k=0;k<4;k++){
		tx=x+next[k][0];
		ty=y+next[k][1];
		if(tx<0||ty<0||tx>=n||ty>=m)
			continue;
		if(book[tx][ty]==0&&s[tx][ty]!='X'){
			book[tx][ty]=1;
			dfs(tx,ty,step+1);
			book[tx][ty]=0;
		}
	}
	return ;
}

int main()
{
	int i,j,k,f;
	while(scanf("%d%d%d",&n,&m,&t)){
		if(n==0&&m==0&&t==0)
			break;
		flag=0;
		for(i=0;i<n;i++){
			scanf("%s",s[i]);
			for(j=0;j<m;j++){
				if(s[i][j]=='S')
					{ sx=i; sy=j;}
				if(s[i][j]=='D')
					{ dx=i; dy=j;}
			}
		}
		memset(book,0,sizeof(book));
		book[sx][sy]=1;
		if(n*m>=t)
			dfs(sx,sy,0);
		if(flag)
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}