poj 1915 Knight Moves(bfs)

题目描述：

在一个n*n 的棋盘上，给一个骑士的起点和一个终点，并给出骑士8种移动方式，求骑士从起点到终点的最小步数.

简单bfs.

CODE:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<map>

#define N 100010
#define Mod 10000007
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
const double EPS = 1e-11;
const double PI = acos ( -1.0 );
typedef long long ll;

const int INF = 1000010;

using namespace std;

int n, x, y, s, e;
bool vis[310][310];

int xx[8] = {2, 1, -1, -2, -2, -1, 1, 2}, yy[8] = {1, 2, 2, 1, -1, -2, -2, -1};

struct moves
{
    int x, y;
    int mos;
} ss;

queue<moves>mp;

int bfs()
{
    memset ( vis, false, sizeof ( vis ) );
    while ( mp.size() )
    {
        mp.pop();
    }
    mp.push ( ss );
    vis[x][y] = true;
    int ans;
    moves t, tt;
    while ( mp.size() )
    {
        tt = mp.front();
        mp.pop();
        if ( tt.x == s && tt.y == e )
            break;
        for ( int i = 0; i < 8; i++ )
        {
            if ( tt.x + xx[i] >= 0 && tt.x + xx[i] < n && tt.y + yy[i] >= 0 && tt.y + yy[i] < n && !vis[tt.x + xx[i]][tt.y + yy[i]] )
            {
                t.x = tt.x + xx[i], t.y = tt.y + yy[i], t.mos = tt.mos + 1;
                mp.push ( t );
                vis[tt.x + xx[i]][tt.y + yy[i]] = true;
            }
        }
    }
    return tt.mos;
}

int main()
{
    int t;
    while ( cin >> t )
    {
        while ( t-- )
        {
            cin >> n >> x >> y >> s >> e;
            ss.x = x;
            ss.y = y;
            ss.mos = 0;
            cout << bfs() << endl;
        }
    }
    return 0;
}