这个题目也是经典的BFS算法，题目的链接如下： http://poj.org/problem?id=1915

这个题目也是经典的BFS算法，题目的链接如下：
http://poj.org/problem?id=1915
感悟：
定义为全局变量，定义结构体，方便进行数之间的步数和位置的操作，直接进行bfs，因为是全局变量，在bfs中进行输出结果

Knight Moves
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 24565 Accepted: 11596
Description

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.

Input

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.
Output

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.
Sample Input

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

5
28
0
Source

题目的大意就是，国际象棋马（骑士）只能走日，然后让你算从起点到重点的最短路径，一般的最短路径什么的都用BFS搜索算法就行了，马走日，八个方向的日都能走，不过，只能走在棋盘中，不能越界。所以要判断重复和越界。

using namespace std;
const int MAXN = 100000 + 10;
int n, sx, sy, ex, ey;
int vis[305][305];
int dirt[8][2] = {-1, -2, -2, -1, -2, 1, -1, 2, 1, -2, 2, -1, 2, 1, 1, 2};
struct node{
    int x, y, step;
};

int judge(int x, int y)
{
    if(x < 0 || x >= n || y < 0 || y >= n)
        return 1 ;
    return vis[x][y];
}

void bfs()
{

    queue<node>q;
    node a, b, next;
    a.x = sx;
    a.y = sy;
    a.step = 0;
    q.push(a);
    vis[sx][sy] = 1;
    while(!q.empty()){
        b = q.front();
        q.pop();
        if(b.x == ex && b.y == ey){
            cout << b.step << endl;
            return;
        }
        for(int i = 0 ; i < 8 ; i++){
        next.x = b.x + dirt[i][0];
        next.y = b.y + dirt[i][1];
        if(judge(next.x, next.y)) continue;
        next.step = b.step + 1;
        q.push(next);
        vis[next.x][next.y] = 1;
        }
    }
}
//input的第一行是样例的个数，
//第二行是棋盘的大小，第三行是起点第四行是终点
int main()
{
    int t ;
    cin >> t;

    while(t--){
    memset(vis , 0 , sizeof(vis));
    cin >> n >> sx >> sy >> ex >> ey ;
    bfs();
    }
    return 0;
}