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…

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: 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…

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 ][] = { {…

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…

http://poj.org/problem?id=2488 题意 : 给你棋盘大小,判断马能否走完棋盘上所有格子,前提是不走已经走过的格子,然后输出时按照字典序排序的第一种路径 思路 : 这个题吧,有点别扭,再加上要用字典序输出,所以就要用一点小技巧了,自己提前将能输出字典序的那个先写到数组里保存,也就是说,搜索方向要进行特殊的排列,而这样的话,只要每次找的时候从第0行第0列开始找,第一个成功走完所有的格子一定是按字典序排列的,因为只要能走完所有的格子,而字典序最小的就是左上角那个格子,所以,…

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…

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 …