HDU1372 跳马问题 BFS

Knight Moves

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1147    Accepted Submission(s): 721

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.".

 

 

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.

 

 

中国象棋跳马问题的变形 ( 国际象棋 ), 宽搜的经典题目 .

无向图的建立 : 以棋盘的 64 个格子做点 , 若马能从一点走一步到达另一点 , 则在这两点之间建立一条边 .

接着用宽搜可以简单得解 .

 

代码如下 :

 

//HDOJ1372 #include<iostream> using namespace std; int h,t,r; void eq(int q[],int d)//入队操作 { q[h++]=d; if (h>200) h=1; } int dq(int q[])//出队操作 { r=q[++t]; q[t]=0; return r; } int BFS(int g[65][65],int s1,int s2)//宽搜 { int k=1,s[65],i,color[65],d[65],q[2000],u; for (i=1;i<=64;++i) { color[i]=0; d[i]=50; } color[s1]=1; d[s1]=0; h=1; t=0; memset(q,0,sizeof(q)); eq(q,s1); while (h-t!=1) { u=dq(q); for (i=1;i<=64;++i) if (g[u][i]==1) { if (color[i]==0) { color[i]=1; d[i]=d[u]+1; if (i==s2) return d[i]; eq(q,i); } } color[u]=2; } } int main() { int g[65][65],i,j,x1,x2,y1,y2,s1,s2;char a[6]; memset(g,0,sizeof(g)); for(j=1;j<=8;++j)//建立图 for (i=1;i<=8;++i) { if ((i+1<=8)&&(j+2<=8)) g[8*(i-1)+j][8*i+j+2]=1; if ((i+2<=8)&&(j+1<=8)) g[8*(i-1)+j][8*(i+1)+j+1]=1; if ((i+2<=8)&&(j-1>0)) g[8*(i-1)+j][8*(i+1)+j-1]=1; if ((i+1<=8)&&(j-2>0)) g[8*(i-1)+j][8*i+j-2]=1; if ((i-1>0)&&(j-2>0)) g[8*(i-1)+j][8*(i-2)+j-2]=1; if ((i-2>0)&&(j-1>0)) g[8*(i-1)+j][8*(i-3)+j-1]=1; if ((i-2>0)&&(j+1<=8)) g[8*(i-1)+j][8*(i-3)+j+1]=1; if ((i-1>0)&&(j+2<=8)) g[8*(i-1)+j][8*(i-2)+j+2]=1; } while (gets(a)!=NULL) { x1=a[0]-'a'+1; y1=a[1]-'0'; x2=a[3]-'a'+1; y2=a[4]-'0'; s1=8*(x1-1)+y1; s2=8*(x2-1)+y2; if (s1==s2) printf("To get from %c%c to %c%c takes 0 knight moves./n",a[0],a[1],a[3],a[4]); else printf("To get from %c%c to %c%c takes %d knight moves./n",a[0],a[1],a[3],a[4],BFS(g,s1,s2)); } return 0; }