poj1915Knight Moves BFS

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

题目大意
求骑士从一个点到另一个点的最少步数

输入
第一行一个t代表t组数据。
每组数据第一行有一个整数n代表一个n*n的棋盘
接下来n行每行四个数代表起点和终点的坐标

输出
每行一个数代表最少步数

#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<string>#include<vector>#include<stack>#include<set>#include<queue>#include<map>using namespace std;const int maxn=301;int a[maxn][maxn],p[maxn][maxn],d[8][2]={{1,2},{1,-2},{-1,2},{-1,-2},{2,1},{2,-1},{-2,1},{-2,-1}}; struct node{    int x,y;};queue <node> q;node make(int x,int y){    node a;    a.x=x;    a.y=y;    return a;}int main(){    int i,j,k,n,m;    int sx,sy,tx,ty,t;    //freopen("test.in","r",stdin);    //freopen("test.out","w",stdout);    scanf("%d",&t);    while(t--){        scanf("%d",&n);        for(i=1;i<=n;i++)            for(j=1;j<=n;j++)                a[i][j]=p[i][j]=0;        while(!q.empty())            q.pop();        scanf("%d%d%d%d",&sx,&sy,&tx,&ty);        sx++;        sy++;        tx++;        ty++;        p[sx][sy]=1;        q.push(make(sx,sy));        while(!q.empty()){            node to;            to=q.front();            q.pop();            if(to.x==tx && to.y==ty) break;            for(i=0;i<8;i++){                int u=to.x+d[i][0],v=to.y+d[i][1];                if(!p[u][v] && u>0 && u<=n && v<=n && v>0){                    q.push(make(u,v));                    p[u][v]=1;                    a[u][v]=a[to.x][to.y]+1;                }            }        }        printf("%d\n",a[tx][ty]);    }    return 0;}