Knight Moves

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8831    Accepted Submission(s): 5202

Problem 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 file 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.".
 
e2 e4 //起点 终点
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.

 #include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
int dx[]={,-,,-,,-,,-};
int dy[]={,-,-,,,-,-,};
int vis[][];
struct point
{
int x;
int y;
int t;
}st,tem,nex;
int sx,sy,ex,ey,tim;
char a,b,c,d;
void bfs()
{
queue<point> s;
st.x=sx,st.y=sy;
st.t=;
s.push(st);
memset(vis,,sizeof(vis));
while(!s.empty())
{
tem=s.front();
s.pop(); if(tem.x==ex&&tem.y==ey)
{
tim=tem.t;
return;
}
if(vis[tem.x][tem.y]||tem.x<=||tem.y<=||tem.x>||tem.y>)
continue;
vis[tem.x][tem.y]=;
for(int i=;i<;i++)
{
int nx=tem.x+dx[i];
int ny=tem.y+dy[i];
int nt=tem.t+;
nex.x=nx,nex.y=ny,nex.t=nt;
s.push(nex);
}
}
}
int main()
{
freopen("in.txt","r",stdin);
while(scanf("%c%c %c%c",&a,&b,&c,&d)!=EOF)
{
getchar();
tim=;
sy=a-'a'+,sx=b-'';
ey=c-'a'+,ex=d-'';
bfs();
printf("To get from %c%c to %c%c takes %d knight moves.\n",a,b,c,d,tim);
}
}
 

Knight Moves(BFS,走’日‘字)的更多相关文章

  1. hdu1372 Knight Moves BFS 搜索

    简单BFS题目 主要是读懂题意 和中国的象棋中马的走法一样,走日字型,共八个方向 我最初wa在初始化上了....以后多注意... 代码: #include <iostream> #incl ...

  2. POJ 2243 Knight Moves(BFS)

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

  3. hdu5794 A Simple Chess 容斥+Lucas 从(1,1)开始出发,每一步从(x1,y1)到达(x2,y2)满足(x2−x1)^2+(y2−y1)^2=5, x2>x1,y2>y1; 其实就是走日字。而且是往(n,m)方向走的日字。还有r个障碍物,障碍物不可以到达。求(1,1)到(n,m)的路径条数。

    A Simple Chess Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)To ...

  4. (step4.2.1) hdu 1372(Knight Moves——BFS)

    解题思路:BFS 1)马的跳跃方向 在国际象棋的棋盘上,一匹马共有8个可能的跳跃方向,如图1所示,按顺时针分别记为1~8,设置一组坐标增量来描述这8个方向: 2)基本过程 设当前点(i,j),方向k, ...

  5. UVA 439 Knight Moves(BFS)

    Knight Moves option=com_onlinejudge&Itemid=8&category=11&page=show_problem&problem=3 ...

  6. HDU 1372 Knight Moves(BFS)

    题目链接 Problem Description A friend of you is doing research on the Traveling Knight Problem (TKP) whe ...

  7. POJ 1915 Knight Moves(BFS+STL)

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

  8. HDU1372:Knight Moves(BFS)

    Knight Moves Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total ...

  9. POJ-1915 Knight Moves (BFS)

    Knight Moves Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 26952   Accepted: 12721 De ...

随机推荐

  1. JavaWeb学习笔记--2.jsp指令元素、动作元素、内置对象

    目录: 一.指令 1.1 page指令 :page指令有13种属性 1.2 include指令:include指令有1种属性,为file.(注意<%@ include%>和<jsp: ...

  2. 【测试技术】ant里面mapper的详细用法

    ant里面mapper标签是和fileset配合使用的,目的就是把fileset取出的文件名转成指定的样式.其实看懂官方文档后,感觉真心没啥好写的.但是还是写一下把. 1.<mapper typ ...

  3. android AndroidManifest.xml 多个android.intent.action.MAIN (

    可以的 ,一个程序是可以有多个入口的.如果你设置两个,并且category都是android.intent.category.LAUNCHER,那么你就 会发现你的手机中就会出现两个快捷方式,也就是两 ...

  4. haporxy 负载elasticsearch

    <pre name="code" class="html">-bash-4.1# cat /etc/haproxy/haproxy.cfg glob ...

  5. NLS_LANG SIMPLIFIED CHINESE_CHINA.AL32UTF8 和american_america.AL32UTF8

    oadb01:/home/oracle> echo $NLS_LANG SIMPLIFIED CHINESE_CHINA.AL32UTF8 oadb01:/home/oracle> sql ...

  6. 提高效率的便签By番茄时间管理 win7标签,小功能,大作用

    今日待办 把一些重要的事情,列入其中. 着重处理. 活动清单 罗列一些最近需要做的事情,不一定按照紧急重要的程度. 把活动清单中的事情,按照实际情况,安排到今日待办当中. 还有一个我喜欢的'头脑风暴' ...

  7. 转:MVC分页

    原文地址:http://www.cnblogs.com/iamlilinfeng/p/4075292.html 分页总是搞得我很烦,也是因为刚接触,貌似有好多插件,之前在用一个,可是后来发现一翻页原来 ...

  8. 【转】repo 的一些用法和理解-不错

    原文网址:http://blog.csdn.net/yasin_lee/article/details/5975068 repo的用法(zz) 注:repo只是google用Python脚本写的调用g ...

  9. VMdomainXml

    1,One,Euc,Ostack 虚拟磁盘镜像制作方法[Windows,Linux,类linux OS](1,基于ios部署系统生成img,2基于vm xml定义部署系统生成img qcow2) 如需 ...

  10. MySQL的备份和还原

    MySQL的备份和还原 备份:副本    RAID1,RAID10:保证硬件损坏而不会业务中止:        DROP TABLE mydb.tb1; 备份类型:        热备份.温备份和冷备 ...