UVA439 骑士的移动 Knight Moves

比较简单的广搜基础题，用来找无权图最短路

#include 

#define fi first
#define se second
#define pb push_back
#define mk make_pair
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()

using namespace std;

typedef long long ll;
typedef vector vi;
typedef pair pa;

int xs, ys, xe, ye;
int dx[8] = {-2, -2, -1, 1, 2, 2, 1, -1};
int dy[8] = {-1, 1, 2, 2, 1, -1, -2, -2};

int bfs() {
    queue q;
    q.push(mk(xs, ys));
    int vis[10][10];
    fill((int*) vis, (int*) vis + 10 * 10, -1);
    vis[xs][ys] = 0;
    while (!q.empty()) {
        pa now = q.front();
        q.pop();
        if (now == mk(xe, ye)) return vis[xe][ye];
        for (int i = 0; i < 8; i++) {
            int x = now.fi + dx[i];
            int y = now.se + dy[i];
            if (x < 0 || x >= 8 || y < 0 || y >= 8) continue;
            if (vis[x][y] != -1) continue;
            q.push(mk(x, y));
            vis[x][y] = vis[now.fi][now.se] + 1;
        }
    }
}

int main() {
    string s1, s2;
    while (cin >> s1 >> s2) {
        xs = s1[0] - 'a'; ys = s1[1] - '1';
        xe = s2[0] - 'a'; ye = s2[1] - '1';
        cout << "To get from " << s1 << " to " << s2 << " takes " << bfs() << " knight moves.n";
    }
    return 0;
}