hoj1440Knight Moves
/*
题目：
象棋中马如何走到指定地点
分析：
BFS题，分8个方向搜就行
具体的图型可以看看poj1915题的8个方向，理解BFS后很容易写出
*/
#include <iostream>
#include <cstring>
#include <queue>
#include <cstdio>
using namespace std;
#define X 310
int sx,sy,ex,ey,n;
bool visit[X][X];
struct node
{
int x,y,step;
};
bool check(int x,int y)
{
if(x<1||y<1||x>n||y>n)
return false;
return true;
}
void bfs()
{
++sx;
++sy;
++ex;
++ey;
queue<node> q;
node a;
node b;
a.step = 0;
a.x = sx;
a.y = sy;
visit[sx][sy] = true;
q.push(a);
int x,y,s;
while(!q.empty())
{
b = q.front();
q.pop();
sx = b.x;
sy = b.y;
s = b.step;
if(sx==ex&&sy==ey)
{
cout<<s<<endl;
return;
}
x = sx-2;
y = sy-1;
if(check(x,y)&&!visit[x][y])
{
visit[x][y] = true;
a.step = s+1;
a.x = x;
a.y = y;
q.push(a);
}
x = sx-2;
y = sy+1;
if(check(x,y)&&!visit[x][y])
{
visit[x][y] = true;
a.step = s+1;
a.x = x;
a.y = y;
q.push(a);
}
x = sx-1;
y = sy-2;
if(check(x,y)&&!visit[x][y])
{
visit[x][y] = true;
a.step = s+1;
a.x = x;
a.y = y;
q.push(a);
}
x = sx-1;
y = sy+2;
if(check(x,y)&&!visit[x][y])
{
visit[x][y] = true;
a.step = s+1;
a.x = x;
a.y = y;
q.push(a);
}
x = sx+1;
y = sy-2;
if(check(x,y)&&!visit[x][y])
{
visit[x][y] = true;
a.step = s+1;
a.x = x;
a.y = y;
q.push(a);
}
x = sx+1;
y = sy+2;
if(check(x,y)&&!visit[x][y])
{
visit[x][y] = true;
a.step = s+1;
a.x = x;
a.y = y;
q.push(a);
}
x = sx+2;
y = sy-1;
if(check(x,y)&&!visit[x][y])
{
visit[x][y] = true;
a.step = s+1;
a.x = x;
a.y = y;
q.push(a);
}
x = sx+2;
y = sy+1;
if(check(x,y)&&!visit[x][y])
{
visit[x][y] = true;
a.step = s+1;
a.x = x;
a.y = y;
q.push(a);
}
}
}
int main()
{
freopen("sum.in","r",stdin);
freopen("sum.out","w",stdout);
int t;
cin>>t;
while(t--)
{
memset(visit,false,sizeof(visit));
scanf("%d%d%d%d%d",&n,&sx,&sy,&ex,&ey);
bfs();
}
return 0;
}