POJ1915 Knight Moves-蒲公英云

POJ1915 Knight Moves

#include <iostream> //结果正确，提交AC
#include <cstdio>   //统计最小步数（故用BFS，直接对应最短路）
#include <queue>
#include <cstring>
using namespace std;
const int MAXN = 100000 + 10;
int n, sx, sy, ex, ey;
int vis[310][310];
int dir[8][2] = {1, -2, 2, -1, 2, 1, 1, 2,-1, -2, -2, -1, -2, 1, -1, 2};   //结果与 此遍历顺序无关
struct node{
    int x;
    int y;
    int step;
};
int judge(int x, int y)
{
    if(vis[x][y]==1 ||x < 0 || x >= n || y < 0 || y >= n)
        return 1 ;
    return 0;
}
void bfs()
{
    queue<node>Q;
    node  now, next;
    now.x = sx;
    now.y = sy;
    now.step = 0;
    Q.push(now);
    vis[sx][sy] = 1;
    while(!Q.empty()){
        now = Q.front();
        Q.pop();
        if(now.x == ex && now.y == ey){
            cout << now.step << endl;
            return;
        }
        for(int i = 0 ; i < 8 ; i++){
            next.x = now.x + dir[i][0];
            next.y = now.y + dir[i][1];
            if( judge(next.x, next.y) )
                continue;
            next.step = now.step + 1;
            Q.push(next);
            vis[next.x][next.y] = 1;
        }
    }
}
int main()
{
    int T ;
    cin >> T;
    while(T--){
        memset(vis,0,sizeof(vis));//每用一次都要置零
        scanf("%d%d%d%d%d",&n,&sx,&sy,&ex,&ey);
        bfs();
    }
    return 0;
}
//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