Tempter of the Bone(奇偶剪枝)の反面教材-优快云博客

本文描述了一个关于迷宫逃脱的问题，迷宫为N*M大小，包含墙壁、起点、终点及可通行区域。需要通过算法帮助一只小狗在限定时间内找到出口。文章提供了使用广度优先搜索(BFS)的尝试性解决方案，并指出了该方法存在的问题。

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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

错误（注意是错误）代码：

#include<iostream>
#include<stdio.h>
#include<queue>
#include<cstring>
#include<stack>
#include<vector>
#include<cmath>


using namespace std;

char board[15][15];
bool book[15][15];
int xs,ys,xd,yd;
int ans;

struct data{
	int x,y,TP;
};
int FX[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};
int N,M,T;

int BFS(){
	queue<data> Q;
	ans=T-abs(xd-xs)-abs(yd-ys); 
	if(ans<0||ans&1) return 0;
	data head;
	head.x = xs,head.y = ys,head.TP = 0;
	Q.push(head);
	while(!Q.empty()){
		data TT = Q.front();
		if(TT.x == xd && TT.y == yd){
			if(TT.TP == T)return 1;
		}
		Q.pop();
		for(int i=0 ; i<4 ; i++){
			int xt = TT.x+FX[i][0];
			int yt = TT.y+FX[i][1];
			if(xt<0 || xt>=N || yt<0 || yt>=M)continue;
			if(book[xt][yt] || board[xt][yt] == 'X')continue;
			if(xt!=xd||yt!=yd)book[xt][yt] = true;
			data mid;
			mid.x = xt,mid.y = yt,mid.TP = TT.TP+1;
			Q.push(mid);
		}
	}
	return 0;
}

int main(){
	while(scanf("%d %d %d",&N,&M,&T) && N|M|T){
		memset(book,false,sizeof(book));
		for(int i=0 ; i<N ; i++){
			scanf("%s",&board[i]);
		}
		int flag = 0;
		for(int i=0 ; i<N ; i++){
			for(int i1=0 ; i1<M ; i1++){
				if(board[i][i1] == 'S'){
					xs = i,ys = i1;
					flag++;
				}
				else if(board[i][i1] == 'D'){
					xd = i,yd = i1;
					flag++;
				}
				if(flag == 2)break;
			}	
			if(flag == 2)break;
		}
		book[xs][ys] == true;
		if(BFS())printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}

这是用BFS写的，讲真，这题应该用深搜，不过因为觉的BFS省时间我开始想都没想直接用的bfs。

这题深搜很简单，所以我就不放代码了（别打我，我承认我一不小心把代码删了，懒的再写了(╯﹏╰)）。