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POJ Knight Moves 2243 x
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.
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 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.
Source
Ulm Local 1996
1)37行是很重要的！好好想想为什么
2)输入的格式get！
代码：
1 #include<cstdio> 2 #include<queue> 3 #include<algorithm> 4 #include<cstring> 5 6 using namespace std; 7 8 char s[5]; 9 bool v[9][9];10 int ex,ey,sx,sy,ans;11 int dx[8]={2,1,-1,-2,-2,-1,1,2};12 int dy[8]={-1,-2,-2,-1,1,2,2,1};//八个方向 13 14 struct node15 {16 int x,y,step;17 }cur,nxt;18 19 queue<node>q;20 21 void bfs()22 {23 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 24 {25 printf("To get from %c%d to %c%d takes %d knight moves.\n",char(ex+‘a‘-1),ey,char(sx+‘a‘-1),sy,0);26 return;//格式 27 }28 while(!q.empty()) q.pop(); // 多组数据初始化 29 memset(v,0,sizeof(v)); // 同上 30 cur.x=ex,cur.y=ey; cur.step=0; //起点 31 v[ex][ey]=true; //不要漏了标记起点 32 q.push(cur);33 while(!q.empty())34 {35 cur=q.front(); 36 q.pop(); //不要漏了当前出队 37 //v[cur.x][cur.y]=false; 出队，清除标记，是否需要？不需要，为什么？ 38 for(int i=0;i<8;i++) //八方位搜索 39 {40 int xx=cur.x+dx[i],yy=cur.y+dy[i]; 41 if(xx>0&&xx<=8&&yy>0&&yy<=8&&!v[xx][yy])42 {43 if(xx==sx&&yy==sy) //找到了，第一个找到的一定就是最近的，why？ 44 {45 printf("To get from %c%d to %c%d takes %d knight moves.\n",char(ex+‘a‘-1),ey,char(sx+‘a‘-1),sy,cur.step+1);46 return ;47 }48 nxt.x=xx, nxt.y=yy; nxt.step=cur.step+1;49 v[nxt.x][nxt.y]=true;50 q.push(nxt); //扩展出的状态入队 51 }52 }53 }54 }55 56 int main()57 {58 while(scanf("%s",s)!=EOF) //注意输入，scanf读到空格 59 {60 ex=s[0]-‘a‘+1; ey=s[1]-‘0‘;61 scanf("%s",s);62 sx=s[0]-‘a‘+1; sy=s[1]-‘0‘;63 bfs();64 }65 }
POJ Knight Moves 2243 x
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