[POJ]1915 Knight Moves

Knight Moves
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 27202 Accepted: 12858
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

TUD Programming Contest 2001, Darmstadt, Germany

简单的bfs
提交了四五次WA
最后发现原因是棋盘从(0,0)开始的 。。
直接把所有++的话n会出问题..

发现了一种记录步数的新方法，在结构体里写step记录，不用在print里把pre遍历一边再计数。

//Writer:GhostCai && His Yellow Duck

#include<iostream>
#include<cstring>
#include<queue> 
#define MAXN 4000
using namespace std;

int t,n,sx,sy,aimx,aimy;
bool vis[MAXN][MAXN];

struct point{
    int x,y,step;
}node,r;

int dx[8]={1,2,2,1,-1,-2,-2,-1};
int dy[8]={2,1,-1,-2,-2,-1,1,2};
point pre[MAXN][MAXN];
bool flag;


void bfs(int x,int y){
    if(x==aimx&&y==aimy){
        cout<<0<<endl;
        return;
    }
    vis[x][y]=1;
    node.x = 0;
    node.y = 0;
    node.step = 0;
    pre[x][y]=node;
    node.x = x;
    node.y = y;
    queue<point> Q;
    Q.push(node) ;
    while(!Q.empty() && !flag){
        r=Q.front() ;
        Q.pop() ;
        int nx,ny;
        for(int i=0;i<=7;i++){
            nx=r.x + dx[i];
            ny=r.y + dy[i];
            if(nx<0||nx>=n||ny<0||ny>=n) continue;
            if(vis[nx][ny]) continue;
            vis[nx][ny]=1;
            pre[nx][ny]=r;
            node.x = nx;
            node.y = ny;
            node.step = r.step + 1;
            Q.push(node);
            if(nx==aimx&&ny==aimy){
                cout<<node.step<<endl ;
                flag=1;
            } 
        }
    }
}

int main(){
    cin>>t;
    while(t--){
        memset(vis,0,sizeof(vis));
        flag=0;
        cin>>n;
        cin>>sx>>sy;
        cin>>aimx>>aimy;
//      n++;
//      sx++;
//      sy++;
//      aimx++;
//      aimy++;
        bfs(sx,sy);
    }
    return 0;
}