e2 e4 a1 b2 b2 c3 a1 h8 a1 h7 h8 a1 b1 c3 f6 f6
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.
<span style="font-size:24px;">#include<iostream>
#include<cstring>
#include<cstdlib>
#include<queue>
using namespace std;
struct node
{
int x,y,num;
};
int a1,a2,b1,b2;
char c1,c2;
int move[8][2]={-2,1,-2,-1,-1,2,-1,-2,1,-2,1,2,2,1,2,-1},v[9][9];
void bfs(int i,int j)
{
node now,temp;
queue<node>q;
now.x=i;
now.y=j;
now.num=0;
memset(v,0,sizeof(v));
q.push(now);
v[now.x][now.y]=1;
while(!q.empty())
{
now=q.front();
q.pop();
if(now.x==b2&&now.y==a2)
{
printf("To get from %c%d to %c%d takes %d knight moves.\n",c1,a1,c2,a2,now.num);
return ;
}
for(int t=0;t<8;t++)
{
temp.x=now.x+move[t][0];
temp.y=now.y+move[t][1];
if(temp.x>0&&temp.x<9&&temp.y>0&&temp.y<9&&!v[temp.x][temp.y])
{
v[temp.x][temp.y]=1;
temp.num=now.num+1;
q.push(temp);
}
}
}
}
int main()
{
while(cin>>c1>>a1>>c2>>a2)
{
b1=c1-'a'+1;
b2=c2-'a'+1;
bfs(b1,a1);
}
return 0;
}</span>杭电 1372 Knight Moves（广搜模板题）,布布扣,bubuko.com
原文地址：http://blog.csdn.net/u012766950/article/details/38119617
