int direct[8][2]={2,1,2,-1,-2,1,-2,-1,1,2,-1,2,1,-2,-1,-2}; //落脚点向量
int judge[4][2]={1,0,-1,0,0,1,0,-1}; //卡位向量
这里说明一下,在使用的时候可以不用太麻烦,我特意将两个向量对应了起来,如
for(i=0;i<4;i++)
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{
if(chess[pos.x+judge[i][0]][pos.y+judge[i][1]]!=1) //是否卡位
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{
int x=pos.x+direct[i*2][0];
int y=pos.y+direct[i*2][1];
if(x<=p&&x>=1&&y<=q&&y>=1&&visited[x][y]!=1&&chess[x][y]!=1) //该点能否落脚
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{
visited[x][y]=1;
temp.x=x;
temp.y=y;
temp.step=pos.step+1;
que.push(temp);
}
x=pos.x+direct[i*2+1][0];
y=pos.y+direct[i*2+1][1];
if(x<=p&&x>=1&&y<=q&&y>=1&&visited[x][y]!=1&&chess[x][y]!=1) //该点能否落脚
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{
visited[x][y]=1;
temp.x=x;
temp.y=y;
temp.step=pos.step+1;
que.push(temp);
}
}
}
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