POJ 2488 A Knight's Journey

http://poj.org/problem?id=2488

　　好久没写回溯，题目要求骑士遍历全图，不能实现就输出“impossible”，首先遍历的

方向要选好，字典序，其次判断遍历了全图的条件就是走了p*q-1步。初始点选取A1即可。

 

/*Accepted    168K    16MS    C++    1267B    2012-07-23 15:35:53*/
#include<cstdio>
#include<cstring>
#include<cstdlib>
const int MAXN = 30;
const int dx[] = {-2, -2, -1, -1, 1, 1, 2, 2};
const int dy[] = {-1, 1, -2, 2, -2, 2, -1, 1};
bool vis[MAXN][MAXN];
int p, q, st, path[MAXN][2];

bool dfs(int x, int y)
{
    int nx, ny, i;
    if( st == p * q - 1) return true;
    for( i = 0; i < 8; i ++)
    {
        nx = x + dx[i];
        ny = y + dy[i];
        if( !vis[nx][ny] && nx >= 0 && nx < q && ny >= 0 && ny < p)
        {
            vis[nx][ny] = true;
            ++ st;
            path[st][0] = nx, path[st][1] = ny;
            if( dfs(nx, ny)) return true;
            -- st;
            vis[nx][ny] = false;
        }
    }
    return false;
}

int main()
{
    int T, cas, i;
    scanf( "%d", &T);
    for( cas = 1; cas <= T; cas ++)
    {
        scanf( "%d%d", &p, &q);
        st = 0;
        memset(vis, false, sizeof vis);
        vis[0][0] = true;
        path[0][1] = path[0][0] = 0;
        printf( "Scenario #%d:\n", cas);
        if( dfs(0, 0))
        {
            for( i = 0; i <= st; i ++)
                printf( "%c%c", 'A' + path[i][0], '1' + path[i][1]);
        }
        else printf( "impossible");
        printf( "\n\n");
    }
    return 0;
}