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…

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: 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…

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: 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…

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…

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…

Time limit1000 ms Memory limit65536 kB 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 a…

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:http://poj.org/problem?id=2488 题目大意:可以从任意点开始,只要能走完棋盘所有点,并要求字典序最小,不可能的话就impossible: 思路:dfs+回溯,因为字典序最小,如果可以的话,肯定是从(1,1)开始的.然后递归搜索该点的所有方向,不能满足就回溯,直到找到能满足的,或者一直找不到. 代码+注释: #include<iostream> #include<cstdio> #include<cstring> #include<…