题目链接:http://poj.org/problem?id=2243 启发式搜索:启发式搜索就是在状态空间中的搜索对每一个搜索的位置进行评估,得到最好的位置,再从这个位置进行搜索直到目标.这样可以省略大量无畏的搜索路径,提到了效率.在启发式搜索中,对位置的估价是十分重要的.采用了不同的估价可以有不同的效果. 估价函数:从当前节点移动到目标节点的预估费用:这个估计就是启发式的.在寻路问题和迷宫问题中,我们通常用曼哈顿(manhattan)估价函数(下文有介绍)预估费用. A*算法与BFS:可以这…

POJ 2243 Knight Moves A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks t…

Knight Moves Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13222   Accepted: 7418 Description A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that v…

POJ 1915 Knight Moves Knight Moves   Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 29912   Accepted: 14058 Description Background Mr Somurolov, fabulous chess-gamer indeed, asserts that no one else but him can move knights from one pos…

1. 图 定义:图(Graph)是由顶点的有穷非空集合和顶点之间边的集合组成,通常表示为:G(V,E),其中,G表示一个图,V是图G中顶点的集合,E是图G中边的集合. 简单点的说:图由节点和边组成.一个节点可能与众多节点直接相连,这些节点被称为邻居. 如二叉树就为一个简单的图: 更加详细的信息可参见:https://www.cnblogs.com/polly333/p/4760275.html 2. 算法 1). 广度优先搜索: 广度优先搜索算法(Breadth First Search,BSF…

1.链接地址: http://bailian.openjudge.cn/practice/1915 http://poj.org/problem?id=1915 2.题目: 总Time Limit: 1000ms Memory Limit: 65536kB Description BackgroundMr Somurolov, fabulous chess-gamer indeed, asserts that no one else but him can move knights from o…

 Knight Moves Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 20913   Accepted: 9702 Description Background Mr Somurolov, fabulous chess-gamer indeed, asserts that no one else but him can move knights from one position to another so fa…

题目: A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks that the most diffi…

题意: 0到N的数轴上,每次可以选择移动到x-1,x+1,2*x,问从n移动到k的最少步数. 思路: 同时遍历三种可能并记忆化入队即可. Tips: n大于等于k时最短步数为n-k. 在移动的过程中可能会越界.重复访问. poj不支持<bits/stdc++.h>和基于范围的for循环. #include <iostream> #include <queue> using namespace std; const int M=110000; struct P{int x…

题目链接:http://poj.org/problem?id=2243 Knight Moves Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14500   Accepted: 8108 Description A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortes…

题 Description A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks that the…

Knight Moves Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13974   Accepted: 7797 Description A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that v…

题目链接 http://poj.org/problem?id=2243 题意 输入8*8国际象棋棋盘上的两颗棋子(a~h表示列,1~8表示行),求马从一颗棋子跳到另一颗棋子需要的最短路径. 思路 使用bfs求解,注意国际象棋中马的走法. 代码 #include <iostream> #include <cstring> #include <cstdio> #include <string> #include <queue> using names…

转载请注明出处:viewmode=contents">http://blog.csdn.net/u012860063?viewmode=contents 题目链接: POJ:http://poj.org/problem?id=2243 HDU: pid=1372">http://acm.hdu.edu.cn/showproblem.php? pid=1372 Problem Description A friend of you is doing research on t…

题目大意:国际象棋给你一个起点和一个终点,按骑士的走法,从起点到终点的最少移动多少次. 求最少明显用bfs,下面给出三种搜索算法程序: // BFS #include<cstdio> #include<queue> #include<cstring> using namespace std; ; int r1,c1,r2,c2; struct Node { int r,c; Node(int r,int c):r(r),c(c){} }; int vis[maxn][m…

Knight Moves Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14125    Accepted Submission(s): 8269 Problem Description A friend of you is doing research on the Traveling Knight Problem (TKP) whe…

# 10028. 「一本通 1.4 例 3」Knight Moves [题目描述] 编写一个程序,计算一个骑士从棋盘上的一个格子到另一个格子所需的最小步数.骑士一步可以移动到的位置由下图给出. [算法] 双向bfs优先扩展节点数少的队列.什么破东西速度没快多少啊.... [代码] #include <stdio.h> #include <queue> #define P pair<int,int> #define ff first #define ss second u…

A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks that the most difficult…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7661    Accepted Submission(s): 4567 Problem Description A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to…

最近在学习广搜  这道题同样是一道简单广搜题=0= 题意:(百度复制粘贴0.0) 题意:给出骑士的骑士位置和目标位置,计算骑士要走多少步 思路:首先要做这道题必须要理解国际象棋中骑士的走法,国际象棋中,骑士是在一个3*2的格子中进行对角线移动,通过画图很容易就知道骑士最多可以朝八个方向移动,那么就朝8个方向进行BFS即可 //细节注意: 1.输入开始结束坐标时需要保存  所以我用了全局变量  因为输出还需要他们=0=: 2.依旧是用bool类型进行map保存,防止内存超限, 0表示还未走过, 1…

Knight Moves Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10542    Accepted Submission(s): 6211 Problem Description A friend of you is doing research on the Traveling Knight Problem (TKP) wh…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1372 Knight Moves Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10372    Accepted Submission(s): 6105 Problem Description A friend of you is doin…

简单搜索,我这里用的是dfs,由于棋盘只有8x8这么大,于是想到dfs应该可以过,后来由于边界的问题,TLE了,改了边界才AC. 这道题的收获就是知道了有些时候dfs没有特定的边界的时候要自己设置一个合适的边界.这题推导可知,任意两点之间马踩6步之内一定能够到达,6步之内还未搜到说明绝对不是最优结果,果断退出.所以这里的res开始时最小设定为6即可,随着设的res的增大,运行时间越来越多,因为深搜可以有很多的分支,不采取较小的边界的话,可能会浪费很多时间在无用的搜索上,所以需要如此剪枝. 反复提…

这道题实到bfs的题目,很简单,不过搜索的方向变成8个而已,对于不会下象棋的会有点晕. #include <iostream> #include <stdio.h> #include <string.h> #include <queue> using namespace std; ][]; ][] = {{-,-},{-,},{-,},{,},{,},{,-},{,-},{-,-}}; typedef struct { int x,y,count; }nod…

其实手写模拟一个队列也挺简单的,尤其是熟练以后. 尼玛,这题欺负我不懂国际象棋,后来百度了下,国际象棋里骑士的走法就是中国象棋里面的马 所以搜索就有八个方向 对了注意初始化标记数组的时候,不要把起点标记为已走过. 因为测试数据里面有一组 f6 f6,此时样例输出的是0 //#define LOCAL #include <iostream> #include <cstdio> #include <cstring> using namespace std; struct P…

题意:给出8*8的棋盘,给出起点和终点,问最少走几步到达终点. 因为骑士的走法和马的走法是一样的,走日字形(四个象限的横竖的日字形) 另外字母转换成坐标的时候仔细一点(因为这个WA了两次---@_@) #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<queue> #define maxn 105 using namespace s…

  // 题意:输入标准国际象棋棋盘上的两个格子,求马最少需要多少步从起点跳到终点 BFS求最短路: bfs并维护距离状态cnt, vis记录是否访问过 #include<cstdio> #include<cstring> #include<iostream> #include<string> #include<algorithm> #include<queue> using namespace std; int r1, c1, r2…

题意:8*8国际象棋棋盘,求马从起点到终点的最少步数. 编写时犯的错误:1.结构体内没构造.2.bfs函数里返回条件误写成起点.3.主函数里取行标时未注意书中的图. #include<iostream> #include<cstdio> #include<cstring> #include<string> #include<sstream> #include<cctype> #include<cmath> #include…

A Knight's Journey Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 31702   Accepted: 10813 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 …

题目例如以下: Knight Moves  A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks t…