poj 1915 Knight Moves （bfs搜索）

Background 
Mr Somurolov, fabulous chess-gamer indeed, asserts that no one else but him can move knights from one position to another so fast. Can you beat him? 
The Problem 
Your task is to write a program to calculate the minimum number of moves needed for a knight to reach one point from another, so that you have the chance to be faster than Somurolov. 
For people not familiar with chess, the possible knight moves are shown in Figure 1. 

The input begins with the number n of scenarios on a single line by itself. 
Next follow n scenarios. Each scenario consists of three lines containing integer numbers. The first line specifies the length l of a side of the chess board (4 <= l <= 300). The entire board has size l * l. The second and third line contain pair of integers {0, ..., l-1}*{0, ..., l-1} specifying the starting and ending position of the knight on the board. The integers are separated by a single blank. You can assume that the positions are valid positions on the chess board of that scenario.

For each scenario of the input you have to calculate the minimal amount of knight moves which are necessary to move from the starting point to the ending point. If starting point and ending point are equal,distance is zero. The distance must be written on a single line.

3
8
0 0
7 0
100
0 0
30 50
10
1 1
1 1

5
28
0

TUD Programming Contest 2001, Darmstadt, Germany

#include <cstdio>
#include <cstring>
const int maxn=100000;//开始设立的是1000，导致一直wa啊
typedef struct Queue
{
    int x,y,count;
}Queue;
Queue que[maxn];//用数组模拟队列
bool visit[310][310];
int dir[8][2]={{-2,1},{-2,-1},{-1,2},{-1,-2},{1,2},{1,-2},{2,1},{2,-1}};//方向数组
int l,c,d;
void bfs(int x,int y)
{
    memset(visit,false,sizeof(visit));//初始化
    int front=0,rear=0;
    int fx,fy,i;
    que[rear].x=x;//入队
    que[rear].y=y;
    que[rear++].count=0;
    visit[x][y]=true;
    while(front<rear)//队不为空
    {
        Queue q=que[front++];
        if(q.x==c && q.y==d)//搜索到结果
        {
            printf("%d\n",q.count);
            break;
        }
        for(i=0;i<8;i++)
        {
            fx=q.x+dir[i][0];
            fy=q.y+dir[i][1];
            if(fx>=0 && fy >=0 && fx<l && fy <l && !visit[fx][fy])//限制条件，这里也要注意，很多人在这里wa的
            {
                visit[fx][fy]=true;//标记访问
                que[rear].x=fx;//入队
                que[rear].y=fy;
                que[rear++].count=q.count+1;
            }
        }
    }
}
int main()
{
    int t,a,b;
    scanf("%d",&t);
    while(t--)
    {
        memset(que,0,sizeof(que));
        scanf("%d",&l);
        scanf("%d%d",&a,&b);
        scanf("%d%d",&c,&d);
        bfs(a,b);
    }
    return 0;
}

思路比较简单，下面是我wa的代码，第二组数据通不过，我也找不到错误。。。

#include <cstdio>
#include <cstring>
typedef struct Queue
{
    int x,y;
    int count;
}Queue;
const int maxn=1000;
int dir[8][2]={{-2,1},{-2,-1},{-1,2},{-1,-2},{1,2},{1,-2},{2,1},{2,-1}};
bool visit[maxn][maxn];
Queue queue[maxn];
int c,d,l;
void bfs(int x,int y)
{
    int front=0,rear=0;
    int i,fx,fy;
    queue[rear].x=x;
    queue[rear].y=y;
    queue[rear++].count=0;
    visit[x][y]=1;
    while(front<rear)
    {
        Queue q=queue[front++];
        if(q.x==c && q.y==d)
        {
            printf("%d\n",q.count);
            break;
        }
        for(i=0;i<8;i++)
        {
             fx=q.x+dir[i][0];
             fy=q.y+dir[i][1];
            if( fx>= 0 && fy>= 0 && fx <l && fy <l&& !visit[fx][fy])
            {
                visit[fx][fy]=1;
                queue[rear].x=fx;
                queue[rear].y=fy;
                queue[rear++].count=q.count+1;
            }
        }
    }
}
int main()
{
    int t;
    int a,b;
    scanf("%d",&t);
    while(t--)
    {
        memset(queue,0,sizeof(queue));
        //memset(visit,0,sizeof(visit));
        scanf("%d",&l);
        scanf("%d%d",&a,&b);
        scanf("%d%d",&c,&d);
        bfs(a,b);
    }
    return 0;
}