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;

}