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题目链接：
POJ：http://poj.org/problem?id=2243
HDU: http://acm.hdu.edu.cn/showproblem.php?pid=1372
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.
题意：用象棋中跳马的走法，从起点到目标点的最小步数；
代码如下：
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstring>
using namespace std;
#define M 1017
struct node
{
int x, y;
int step;
};
int xx[8] = {1,2,2,1,-1,-2,-2,-1};
int yy[8] = {2,1,-1,-2,-2,-1,1,2};
bool vis[M][M];
int n, ansx, ansy;
queue<node>q;
int BFS(int x, int y)
{
if(x == ansx && y == ansy)
return 0;
int dx, dy, i;
node front, rear;
front.x = x, front.y = y, front.step = 0;
q.push(front);
vis[x][y] = true;
while(!q.empty())
{
front = q.front();
q.pop();
for(i = 0; i < 8; i++)
{
dx = front.x+xx[i];
dy = front.y+yy[i];
if(dx>=1&&dx<=8&&dy>=1&&dy<=8&&!vis[dx][dy])
{
vis[dx][dy] = true;
if(dx == ansx && dy == ansy)
{
return front.step+1;
}
rear.x = dx, rear.y = dy, rear.step = front.step+1;
q.push(rear);
}
}
}
}
int main()
{
char s,e;
int a1,a2;
while(~scanf("%c%d %c%d",&s,&a1,&e,&a2))
{
getchar();
while(!q.empty())
q.pop();
memset(vis,0,sizeof(vis));
int s1 = s-'a'+1;
int e1 = e-'a'+1;
ansx = e1, ansy = a2;
int ans = BFS(s1,a1);
printf("To get from %c%d to %c%d takes %d knight moves.\n",s,a1,e,a2,ans);
}
return 0;
}poj2243&&hdu1372 Knight Moves（BFS）,布布扣,bubuko.com
poj2243&&hdu1372 Knight Moves（BFS）
原文地址：http://blog.csdn.net/u012860063/article/details/37761995
