poj 1915/2243 bfs（马走日）

题意：给定一个n*n的棋盘，一个起点坐标，一个终点坐标。问从起点“走日”到终点的最少步数。走日即象棋里的马走日。(2243题意完全相同，只不过棋盘限制在8*8)

思路：bfs水题。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <cstdlib>
using namespace std;
#define clc(s,t) memset(s,t,sizeof(s))
#define INF 0x3fffffff
#define N 305
int s[N][N],flag[N][N];
int T,n;
int a,b,x,y;
int ori[8][2] = {{-2,-1},{-2,1},{-1,2},{1,2},{2,1},{2,-1},{1,-2},{-1,-2}};
struct node{
    int x,y,t;
    node(int xx,int yy,int tt):x(xx),y(yy),t(tt){};
};
queue<struct node>q;
int check(int x,int y){
    return x>=0&&y>=0&&x<n&&y<n;
}
int bfs(){
    while(!q.empty()){
        struct node now = q.front();
        q.pop();
        if(now.x == x && now.y == y)
            return now.t;
        for(int i = 0;i<8;i++){
            int xx = now.x+ori[i][0];
            int yy = now.y+ori[i][1];
            if(check(xx, yy) && !flag[xx][yy]){
                q.push(node(xx,yy,now.t+1));
                flag[xx][yy] = 1;
            }
        }
    }
    return -1;
}
int main(){
    scanf("%d",&T);
    while(T--){
        while (!q.empty())
            q.pop();
        clc(flag, 0);
        scanf("%d",&n);
        scanf("%d %d %d %d",&a,&b,&x,&y);
        q.push(node(a,b,0));
        flag[a][b] = 1;
        printf("%d\n",bfs());
    }
    return 0;
}