uva 439 - Knight Moves
来源:互联网 发布:桌面壁纸制作软件 编辑:程序博客网 时间:2024/05/20 11:46
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 Specification
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 Specification
For each test case, print one line saying "To get from xx to yy takes n knight moves.".
Sample Input
e2 e4a1 b2b2 c3a1 h8a1 h7h8 a1b1 c3f6 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.
这个题就是开始还以为是一格一格走的,后面才知道原来是按中国象棋里面马的那种走法,那么做法就是从当前位置出发,不断扩张更新,在更新的过程中如果有相同位置情况就比较一下,最后输出:
#include<stdio.h>#include<stdlib.h>#include<string.h>int maze[10][10];//用于存储从起点到所在点的最小步数int dir[8][2]={{-2,-1},{-2,1},{2,-1},{2,1},{1,2},{1,-2},{-1,2},{-1,-2}};//代表八个方向 int x1,x2,y1,y2;//起点和终点坐标 void dfs(int x,int y,int step){ if(x==x2&&y==y2)return; for(int i=0;i<8;i++){ int a=x+dir[i][0],b=y+dir[i][1]; if((a>=0&&a<=7&&b>=0&&b<=7)&&(maze[a][b]==-1||maze[a][b]>step)){ maze[a][b]=step; dfs(a,b,step+1); } } return;} int main(){ char s1[5],s2[5]; while(scanf("%s%s",s1,s2)!=EOF){ memset(maze,-1,sizeof(maze)); x1=s1[0]-97,x2=s2[0]-97,y1=s1[1]-49,y2=s2[1]-49; maze[x1][y1]=0;//起始点得注意 for(int i=0;i<8;i++){//遍历八个方向 int x=x1+dir[i][0], y=y1+dir[i][1]; if(x>=0&&x<=7&&y>=0&&y<=7){//越的就不用再管了 maze[x][y]=1; dfs(x,y,2); } } printf("To get from %s to %s takes %d knight moves.\n",s1,s2,maze[x2][y2]); } return 0;}
- uva 439 knight moves
- uva 439 - Knight Moves
- uva 439 - Knight Moves
- uva 439 - Knight Moves
- uva 439 - Knight Moves
- uva-439 - Knight Moves
- UVa 439 - Knight Moves
- UVa 439 - Knight Moves
- UVA 439 - Knight Moves
- uva 439 Knight Moves
- UVa 439 - Knight Moves
- uva 439 - Knight Moves
- UVa:439 Knight Moves
- uva 439 - Knight Moves
- uva 439 Knight Moves
- UVa 439 - Knight Moves
- uva 439 - Knight Moves
- UVA 439 - Knight Moves
- Hdu 4451 Dressing
- Python读取UTF-16
- POJ 3468 (线段树,区间更新,查询区间)
- Foxmail 7 客户端配置 Exchange2013 邮箱帐户
- COM学习笔记——基础
- uva 439 - Knight Moves
- 用JQuery来完成XML的解析
- struts2 HelloWorld
- oninput,onpropertychange,onchange的用法和区别
- windows环境下Socket编程的几种模式
- chrome不支持12号以下字体的解决方法
- unity3D与网页的交互
- 带复选框的树形控件实例
- 【BCB】CB主程序 启动卡住 运行无响应 报错解决办法