迷宫问题(bfs) POJ - 3984   #include <iostream> #include <queue> #include <stack> #include <cstring> using namespace std; /*广度优先搜索*/ /*将每个未访问过的邻接点进队列,然后出队列,知道到达终点*/ typedef class { public: int x; int y; }coordinate; ][]; //迷宫 ][] = { {…

A Knight's Journey Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 41936   Accepted: 14269 Description Background The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey ar…

Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 25950   Accepted: 8853 Description Background The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey around the world. When…

http://poj.org/problem?id=2488 Description Background The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey around the world. Whenever a knight moves, it is two squares in one directi…

Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 45941   Accepted: 15637 Description Background The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey around the world. Whe…

题目链接. 题目大意: 给定一个矩阵,马的初始位置在(0,0),要求给出一个方案,使马走遍所有的点. 列为数字,行为字母,搜索按字典序. 分析: 用 vis[x][y] 标记是否已经访问.因为要搜索所有的可能,所以没搜索完一次要把vis[x][y]标记为未访问.详情见代码. 用 p[x][y] 表示 (x,y)点的祖先,以便输出路径. dfs搜索即可. #include <iostream> #include <cstdio> #include <string> #in…

题目链接:http://poj.org/problem?id=2488 A Knight's Journey Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 36695   Accepted: 12462 Description Background The knight is getting bored of seeing the same black and white squares again and again…

A Knight's Journey Time Limit: 1000MSMemory Limit: 65536K Total Submissions: 34633Accepted: 11815 Description BackgroundThe knight is getting bored of seeing the same black and white squares again and again and has decided to make a journeyaround the…

A Knight’s Journey Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 34085 Accepted: 11621 Description Background The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey around…

A Knight's Journey Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 35868   Accepted: 12227 Description Background The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey ar…

A Knight's Journey Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 31147   Accepted: 10655 Description Background  The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey …

A Knight's Journey Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 35564 Accepted: 12119 Description Background The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey around…

A Knight's Journey Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 51   Accepted Submission(s) : 17 Problem Description Background The knight is getting bored of seeing the same black and white…

  A Knight's Journey Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 24840   Accepted: 8412 Description Background   The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey…

2015-06-05 问题简述: 有一个 p*q 的棋盘,一个骑士(就是中国象棋里的马)想要走完所有的格子,棋盘横向是 A...Z(其中A开始 p 个),纵向是 1...q. 原题链接:http://acm.tju.edu.cn/toj/showp1702.html 解题思路: DFS:深搜把所有情况都考虑一遍,当骑士走出棋盘或走到原来走过的格子时,旅行失败了:当骑士能一直走下去直到走过的格子数等于 p*q 时,旅行成功. 提交过程中一直有一个问题使这个代码一直WA,最后才发现是 p.q输入反了…

A Knight's Journey Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 66   Accepted Submission(s) : 27 Problem Description Background The knight is getting bored of seeing the same black and white…

A Knight's Journey Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 31195   Accepted: 10668 Description Background The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey ar…

题目链接:http://vjudge.net/contest/view.action?cid=51369#problem/A   (A - Children of the Candy Corn) http://poj.org/problem?id=2488   (A Knight's Journey) (不知道为什么,名字竟然不同哇~~~~还是poj 改名改得好) 题目意思:给出一个p * q 的棋盘,行用阿拉伯数字1,2,...,p-1, p 来表示,列从大写字母'A'开始表示.我这里为了简化…

POJ 2488 -- A Knight's Journey(骑士游历) 题意: 给出一个国际棋盘的大小,判断马能否不重复的走过所有格,并记录下其中按字典序排列的第一种路径. 经典的“骑士游历”问题 输入:第一行,整数n,接下来是n行,每一行为p和q,p为行数,q为列数,p用1...p编号,q用A...Q编号 马的走法:每步棋先横走或直走一格,然后再往外斜走一格;或者先斜走一格,最后再往外横走或竖走一格(即走"日"字).可以越子,没有中国象棋中的"蹩马腿"限制. 解…

A Knight's Journey Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 35342   Accepted: 12051 Description Background  The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey …

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=1157 题目描述 小鼠a与小鼠b身处一个m×n的迷宫中,如图所示.每一个方格表示迷宫中的一个房间.这m×n个房间中有一些房间是封闭的,不允许任何人进入.在迷宫中任何位置均可沿上,下,左,右4个方向进入未封闭的房间.小鼠a位于迷宫的(p,q)方格中,它必须找出一条通向小鼠b所在的(r,s)方格的路.请帮助小鼠a找出所有通向小鼠b的最短道路. 请编程对于给定…

题目: Background The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey around the world. Whenever a knight moves, it is two squares in one direction and one square perpendicular to this…

poj-2488 题意:一个人要走遍一个不大于8*8的国际棋盘,他只能走日字,要输出一条字典序最小的路径 题解: (1)题目上说的"The knight can start and end on any square of the board.",是个坑点,其实要走字典序最小只需从A1开始遍历就行,因为从任意一点开始,只要能遍历完整个地图,那么A1也可以: (2)要使字典序最小,那么遍历顺序一定要注意 int dr[8]={-1,1,-2,2,-2,2,-1,1};int dc[8]=…

简单搜索,在n*n的矩阵中,问从起点是否可以到达终点,有些格子不可走,上下左右四个方向都可以走.(N<=100)1.bfs从起点开始走,直到走到终点或全部遍历过一次就结束.2.dfs要一走到终点就返回,否则4^n会TLE.由于询问"是否可到达终点",就直接递归"是否可以走到点(x,y)点"的函数,也是直到找到终点就结束. 下面附上dfs的代码-- 1 #include<cstdio> 2 #include<cstdlib> 3 #inc…

题目:http://poj.org/problem?id=2488 题意: 给出一个国际棋盘的大小,判断马能否不重复的走过所有格,并记录下其中按字典序排列的第一种路径. #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack> #include<queue> #include<iomanip> #incl…

题目地址:http://poj.org/problem?id=2488 Sample Input 3 1 1 2 3 4 3 Sample Output Scenario #1: A1 Scenario #2: impossible Scenario #3: A1B3C1A2B4C2A3B1C3A4B2C4 题目:给你一个p*q的棋盘,规则布局参考上图.列用字母表示,行用数字表示.这样一个左上角的节点就是(A,1).骑士的棋子走日字形状,可以从任意节点出发,终点也是任意的.问你能不能遍历所有的位…

给一个矩形棋盘,每次走日字,问能否不重复的走完棋盘的每个点,并将路径按字典序输出 *解法:按字典序输出路径,因此方向向量的数组按字典序写顺序,dfs+回溯,注意flag退出递归的判断,并且用pre记录路径 #include <iostream> #include <cstdio> #include <cstring> using namespace std; char a[30][30]; int dx[] = {-2, -2, -1, -1, 1, 1, 2, 2};…

补个很久之前的题解.... 题目链接: http://poj.org/problem?id=2488 题意: 马走"日"字,让你为他设计一条道路,走遍所有格,并输出字典序最小的一条. 分析: dfs~~~ 代码: #include<iostream> #include<cstring> #include<algorithm> using namespace std; typedef pair<int, int>pii; const int…

题目链接:http://poj.org/problem?id=2488 题意: 在国际象棋的题盘上有一个骑士,骑士只能走“日”,即站在某一个位置,它可以往周围八个满足条件的格子上跳跃,现在给你一个p * q的矩形格子,让你找一个跳跃顺序(起点自选),使得这个顺序恰好经过矩阵的每一个格子,且每一个格子仅经过一次,即找一个符合跳跃条件的序列,遍历整个矩形格子.如果有多个,那么就输出字典序最小的. 思路: 貌似可以利用哈密顿通路来解决,但是感觉有点太麻烦,没怎么细想,感觉还是回溯法比较好.首先题目要求…

import java.util.LinkedList; import java.util.Queue; import java.util.Stack; public class BFS { private static final int M=3;//代表行 private static final int N=3;//代表列 int destX=1; int destY=0; //int[][] maze={{0,0,1,1},{0,0,0,1},{1,1,0,1},{0,0,0,0}};/…