Knight Moves
Time Limit:3000MSMemory Limit:0KB64bit IO Format:%lld & %llu
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Description
A friend of you is doing research on theTraveling Knight Problem (TKP)where you are to find the shortest closed tour of knight moves that visits each square of a given set ofnsquares 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 squaresaandbas input and then determines the number of knight moves on a shortest route fromatob.
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 fromxxtoyytakesnknight moves.".
Sample Inpute2 e4a1 b2b2 c3a1 h8a1 h7h8 a1b1 c3f6 f6Sample OutputTo 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.
题意：国际象棋中的“骑士”（我们俗称“马”，以下亦简称“马”）走的步法规则和中国象棋中的马类似，走“日”字格，如下图所示中“马”可走的位置为标号1（左上纵）、2（右上纵）、3（左上横）、4（右上横）、5（左下横）、6（右下横）、7（左下纵）、8（右下纵）。题目用字母a~h表示棋盘横坐标(指示棋盘纵列)，数字1~8表示棋盘纵坐标(指示棋盘横列)，，则字母数字组合表示了棋盘的某个位置，如a1即表示棋盘左上角第一个位置。输入有多组用例，每组用例用上述的字母数字组合给出“马”的起始位置和目标位置，需输出“马”从起始位置到目标位置的最少步数。
思路：懂得了国际象棋的规则，就会发现是一个简单的bfs
#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>#include <set>#include <queue>#include <stack>using namespace std;struct node{int x,y;int step;};int map[110][110];int vis[110][110];int jx[]={2,1,1,2,-1,-2,-1,-2};int jy[]={1,2,-2,-1,2,1,-2,-1};char str1[20],str2[20];int n,m;void bfs(int xx,int yy){int i;struct node t,f;queue<node >q;t.x=xx;t.y=yy;t.step=0;q.push(t);vis[t.x][t.y]=1;while(!q.empty()){t=q.front();q.pop();if(map[t.x][t.y]==2){printf("To get from %s to %s takes %d knight moves.\n",str1,str2,t.step);return ;}for(i=0;i<8;i++){f.x=t.x+jx[i];f.y=t.y+jy[i];if(f.x>=0&&f.x<8&&f.y>=0&&f.y<8&&!vis[f.x][f.y]){vis[f.x][f.y]=1;f.step=t.step+1;q.push(f);}}}}int main(){int s1,s2,t1,t2;while(~scanf("%s %s",str1,str2)){memset(map,0,sizeof(map));memset(vis,0,sizeof(vis));s1=str1[0]-'a';t1=str1[1]-'1';map[s1][t1]=1;s2=str2[0]-'a';t2=str2[1]-'1';map[s2][t2]=2;bfs(s1,t1);}return 0;}
走走停停，不要害怕错过什么，
