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 visits each square of a given set of n squares on a chessboard exactly once. He thinks that the most difficult part of the problem is determining the smallest number of knight moves between two given squares and that, once you have accomplished this, finding the tour would be easy. 
Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part.

Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b.

Input

The input will contain one or more test cases. Each test case consists of one line containing two squares separated by one space. A square is a string consisting of a letter (a-h) representing the column and a digit (1-8) representing the row on the chessboard.

Output

For each test case, print one line saying "To get from xx to yy takes n knight moves.".

Sample Input

e2 e4
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6

Sample Output

To get from e2 to e4 takes 2 knight moves.
To get from a1 to b2 takes 4 knight moves.
To get from b2 to c3 takes 2 knight moves.
To get from a1 to h8 takes 6 knight moves.
To get from a1 to h7 takes 5 knight moves.
To get from h8 to a1 takes 6 knight moves.
To get from b1 to c3 takes 1 knight moves.
To get from f6 to f6 takes 0 knight moves.

Source

 
1)37行是很重要的!好好想想为什么
2)输入的格式get!
 
代码:
 #include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring> using namespace std; char s[];
bool v[][];
int ex,ey,sx,sy,ans;
int dx[]={,,-,-,-,-,,};
int dy[]={-,-,-,-,,,,};//八个方向 struct node
{
int x,y,step;
}cur,nxt; queue<node>q; void bfs()
{
if(ex==sx&&ey==sy) //特判起点等于终点 ,找到后进行输出就一定是最小的
{
printf("To get from %c%d to %c%d takes %d knight moves.\n",char(ex+'a'-),ey,char(sx+'a'-),sy,);
return;//格式
}
while(!q.empty()) q.pop(); // 多组数据初始化
memset(v,,sizeof(v)); // 同上
cur.x=ex,cur.y=ey; cur.step=; //起点
v[ex][ey]=true; //不要漏了标记起点
q.push(cur);
while(!q.empty())
{
cur=q.front();
q.pop(); //不要漏了当前出队
//v[cur.x][cur.y]=false; 出队,清除标记,是否需要?不需要,为什么?
for(int i=;i<;i++) //八方位搜索
{
int xx=cur.x+dx[i],yy=cur.y+dy[i];
if(xx>&&xx<=&&yy>&&yy<=&&!v[xx][yy])
{
if(xx==sx&&yy==sy) //找到了,第一个找到的一定就是最近的,why?
{
printf("To get from %c%d to %c%d takes %d knight moves.\n",char(ex+'a'-),ey,char(sx+'a'-),sy,cur.step+);
return ;
}
nxt.x=xx, nxt.y=yy; nxt.step=cur.step+;
v[nxt.x][nxt.y]=true;
q.push(nxt); //扩展出的状态入队
}
}
}
} int main()
{
while(scanf("%s",s)!=EOF) //注意输入,scanf读到空格
{
ex=s[]-'a'+; ey=s[]-'';
scanf("%s",s);
sx=s[]-'a'+; sy=s[]-'';
bfs();
}
}

POJ Knight Moves 2243 x的更多相关文章

  1. POJ 2243 Knight Moves(BFS)

    POJ 2243 Knight Moves A friend of you is doing research on the Traveling Knight Problem (TKP) where ...

  2. 【POJ 2243】Knight Moves

    题 Description A friend of you is doing research on the Traveling Knight Problem (TKP) where you are ...

  3. POJ 2243 Knight Moves

    Knight Moves Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13222   Accepted: 7418 Des ...

  4. OpenJudge/Poj 1915 Knight Moves

    1.链接地址: http://bailian.openjudge.cn/practice/1915 http://poj.org/problem?id=1915 2.题目: 总Time Limit: ...

  5. POJ 1915 Knight Moves

    POJ 1915 Knight Moves Knight Moves   Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 29 ...

  6. POJ 1915 Knight Moves(BFS+STL)

     Knight Moves Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 20913   Accepted: 9702 ...

  7. HDU 2243 Knight Moves

    题目: A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find th ...

  8. POJ---2243 Knight Moves 使用A*算法的广度优先搜索

    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 ...

  9. poj2243 Knight Moves(BFS)

    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 ...

随机推荐

  1. python string_1

    quote :http://www.runoob.com/python/python-strings.html #coding:utf-8 s1="http://www.jnshu.com/ ...

  2. 20191127 Spring Boot官方文档学习(4.25)

    4.25. Testing Spring Boot提供了许多实用程序和注解,可以在测试应用程序时提供帮助.测试支持由两个模块提供:spring-boot-test包含核心项,spring-boot-t ...

  3. nrm安装与配置

    nrm安装与配置:https://blog.csdn.net/anway12138/article/details/79455224

  4. SVN与Git的优点差异比较

    今天自己还是很有进步的,但是 下午的进度很慢,学习还是得回去,不能在工位进行 在网上看到一篇有关于SVN与Git的区别 复制下来了,以后可以经常看看 一. 集中式vs分布式 1. Subversion ...

  5. [LeetCode] 212. 单词搜索 II

    题目链接:https://leetcode-cn.com/problems/word-search-ii/ 题目描述: 给定一个二维网格 board 和一个字典中的单词列表 words,找出所有同时在 ...

  6. MVCC/分布式事务简介

    之前我们学习了RocksDB,但这还只是一个最基础的存储引擎.如果想把它在生产环境中用起来,还需要解决很多问题: 如何从单机扩展到分布式? 如何实现事务,并对事务进行并发控制? 用户接口能不能高级一点 ...

  7. js面向过程-拖拽

    1.步骤分析: 1.1 获取id 1.2 当鼠标点击时执行的js 1.3当鼠标移动时执行的js 1.4当鼠标放开时执行的js 2.代码实现 <!DOCTYPE html> <html ...

  8. DOM属性和事件

    1-22 DOM属性设置与获取   1.获取属性: getAttribute("attribute"): var p = document.getElementById(" ...

  9. 【React -- 9/100】 抽离顶部导航栏 - [组件复用]

    今天写的页面中需要重复使用到顶部导航栏,所以把顶部导航栏抽离出来 考虑复用组件的健壮性,使用PropTypes校验,可以自定义一个click事件 JSX import React from " ...

  10. js中的Math对象

    绝对值Math.abs()     console.log(Math.abs(-25));     console.log(Math.abs('-25'));//存在隐式转换可以求绝对值     co ...