$des$

实现一个bfs

$sol$

写了一个双向bfs

#include 
     
      using namespace std;#define Rep(i, a, b) for(int i = a; i <= b; i ++)#define gc getchar()inline int read() {    int x = 0; char c = gc;    while(c < '0' || c > '9') c = gc;    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc;    return x;}const int N = 305;const int xd[] = {-1, -2, -2, -1, 1, 2, 2, 1};const int yd[] = {-2, -1, 1, 2, 2, 1, -1, -2};int Lim = 300;int vis[N][N], bel[N][N];int n;int Bfs_time;queue
      
       
         > Q;inline int Work(int sx, int sy, int tx, int ty) {    if(sx == tx && sy == ty) return 0;    memset(vis, 0, sizeof vis);    memset(bel, 0, sizeof bel);    while(!Q.empty()) Q.pop();    Q.push(make_pair(sx, sy));    Q.push(make_pair(tx, ty));    bel[sx][sy] = 1, bel[tx][ty] = 2;    vis[tx][ty] = 1;    while(!Q.empty()) {        pair 
        
          tp = Q.front();        Q.pop();        int px = tp.first, py = tp.second;        Rep(i, 0, 7) {            int nx = px + xd[i], ny = py + yd[i];            if(bel[nx][ny] == bel[px][py] || nx < 0 || nx > Lim || ny < 0 || ny > Lim) continue;            if(vis[nx][ny]) return vis[nx][ny] + vis[px][py];            vis[nx][ny] = vis[px][py] + 1;            bel[nx][ny] = bel[px][py];            Q.push(make_pair(nx, ny));        }    }}int main() {    n = read();    while(n --) {        Lim = read();        cout << Work(read(), read(), read(), read()) << "\n";        }    return 0;}