POJ 1915-Knight Moves

http://poj.org/problem?id=1915

这道题还是求骑士从一个点移动到另一个点距离的问题，不同的是棋盘的规格是由我们输入的，

最大有300*300，数组要开足够大，我把2243的代码修改了下就AC了...好好理解下BFS

#include<iostream>
#include<string.h>
using namespace std;

int x1, y1, x2, y2;
int dist[305][305], qx[90005], qy[90005];
int dx[] = { -1, -2, -2, -1, 1, 2, 2, 1};
int dy[] = { -2, -1, 1, 2, 2, -1, 1, -2};

int main()
{
    int n, m, nx, ny, x, y, rear, front;
    cin >> m;
    while( m --)
    {
        cin >> n;
        cin >> x1 >> y1;
        cin >> x2 >> y2;
        memset( dist, -1, sizeof( dist) );
        dist[x1][y1] = 0;
        front = rear = 0;
        qx[rear] = x1;
        qy[rear] = y1;
        rear ++;
        while( front < rear)
        {
            x = qx[front];
            y = qy[front];
            front ++;
            if( x == x2 && y == y2)
                break;
            for( int i = 0; i < 8; i ++)
            {
                nx = x + dx[i];
                ny = y + dy[i];
                if( dist[nx][ny] < 0 && nx >= 0 && nx < n && ny >= 0 && ny < n)
                {
                    dist[nx][ny] = dist[x][y] + 1;
                    qx[rear] = nx;
                    qy[rear] = ny;
                    rear ++;
                }
            }
        }
        int steps = dist[x2][y2];
        cout << steps << endl;
    }
    return 0;
}

 

今天看了下C ++的STL库，提到了pair，想着去试试。然后写了一遍，STL果然很方便。

#include<iostream>
#include<queue>
#define MAXN 305
using namespace std;
typedef pair< int, int> pii;
int d[MAXN][MAXN], N;
const int dx[] = { 1, 1, 2, 2, -1, -1, -2, -2};
const int dy[] = { 2, -2, 1, -1, 2, -2, 1, -1};
int x1, y1, x2, y2;
void bfs()
{
    queue<pii> q;
    q.push( make_pair( x1, y1) );
    d[x1][y1] = 0;
    while( !q.empty() )
    {
        pii u = q.front();  q.pop();
        int x = u.first, y = u.second;
        if( x == x2 && y == y2) break;
        for( int i = 0; i < 8; i ++)
        {
            int nx = x + dx[i];
            int ny = y + dy[i];
            if( d[nx][ny] < 0 && nx >= 0 && nx < N && ny >= 0 && ny < N)
            {
                d[nx][ny] = d[x][y] + 1;
                q.push( make_pair( nx, ny));
            }
        }
    }
}

int main()
{
    int cas;
    cin >> cas;
    while( cas --)
    {
        cin >> N;
        cin >> x1 >> y1;
        cin >> x2 >> y2;
        memset( d, -1, sizeof d);
        bfs();
        cout << d[x2][y2] << endl;
    }
    return 0;
}