UVa 439 Knight Moves

　　给一个8*8的棋盘，给出骑士的起点和终点，求出从起点到终点的走的最小步数。这个也是ZOJ 的1091题。

　　用bfs解决，一个vis数组保存是否访问过，dis数组保存从起点到该点的移动次数。代码如下：

#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;

queue<int> q;
int d[][2] = {{2, 1}, {1, 2}, {-1, 2}, {-2, 1}, {-2, -1}, {-1, -2}, {1,-2}, {2, -1}};
int vis[8][8], dis[8][8];
int end_x, end_y;

int bfs(int x, int y)
{
    vis[x][y] = 1;
    dis[x][y] = 0;
    while(!q.empty())   q.pop();
    int t;
    t = x * 8 + y;
    q.push(t);
    while(!q.empty())
    {
        t = q.front();
        q.pop();
        x = t / 8;
        y = t % 8;
        if(x == end_x && y == end_y)   return dis[x][y];
        for(int i = 0; i < 8; i++)
        {
            int nx = x+d[i][0], ny = y+d[i][1];
            if(nx >= 0 && nx < 8 && ny >= 0 && ny < 8 && !vis[nx][ny])
            {
                vis[nx][ny] = 1;
                dis[nx][ny] = dis[x][y] + 1;
                t = nx * 8 + ny;
                q.push(t);
            }
        }
    }
    return -1;
}

int main()
{
#ifdef LOCAL
    freopen("in", "r", stdin);
#endif
    char start[5], end[5];
    while(scanf("%s%s", start, end) != EOF)
    {
        int x, y;
        x = start[1] - '1';
        y = start[0] - 'a';
        end_x = end[1] - '1';
        end_y = end[0] - 'a';
        memset(vis, 0, sizeof(vis));
        memset(dis, 0, sizeof(dis));
        int ans = bfs(x, y);
        printf("To get from %s to %s takes %d knight moves.\n", start, end, ans);
    }
    return 0;
}