URAL 1016 Cube on the Walk

URAL_1016

    由于立方体位于同一个点时可能有不同的形态，从而有不同的结果，因此我们可以根据立方体的形态将棋盘上一个点拆成若干个点，然后做最短路即可。当然，由于状态比较复杂，可以用哈希表映射出每个点的存储位置。

#include<stdio.h>
#include<string.h>
#define MAXD 1000010
#define Q 1000000
#define HASH 1000003
#define INF 0x3f3f3f3f
int dx[] = {-1, 1, 0, 0}, dy[] = {0, 0, 1, -1};
int ch[][6] = {{0, 1, 5, 2, 3, 4}, {0, 1, 3, 4, 5, 2}, {2, 4, 1, 3, 0, 5}, {4, 2, 0, 3, 1, 5}};
char b[10];
int sx, sy, tx, ty, qx[MAXD], qy[MAXD], st[MAXD][6], v[MAXD], where[MAXD];
struct Hashmap
{
    int f[MAXD], head[HASH], next[MAXD], e, state[MAXD][6], x[MAXD], y[MAXD], p[MAXD];
    void init()
    {
        memset(head, -1, sizeof(head));
        e = 1;
    }
    int hash(int s)
    {
        int i, h, seed = 131;
        h = qx[s] * seed + qy[s];
        for(i = 0; i < 6; i ++)
            h = h * seed + st[s][i];
        return (h & 0x7fffffff) % HASH;
    }
    int push(int s, int fa)
    {
        int i, h = hash(s);
        for(i = head[h]; i != -1; i = next[i])
            if(x[i] == qx[s] && y[i] == qy[s] && memcmp(st[s], state[i], sizeof(state[i])) == 0)
                break;
        if(i == -1)
        {
            x[e] = qx[s], y[e] = qy[s], memcpy(state[e], st[s], sizeof(st[s])), f[e] = v[s];
            p[e] = fa;
            next[e] = head[h], head[h] = e;
            return e ++;
        }
        if(v[s] < f[i])
        {
            f[i] = v[s], p[i] = fa;
            return i;
        }
        return 0;
    }
}hm;
void print(int cur)
{
    if(hm.p[cur] != -1)
        print(hm.p[cur]);
    printf(" %c%c", hm.x[cur] + 'a', hm.y[cur] + '1');
}
void solve()
{
    int i, j, k, front, rear, x, y, newx, newy, ans = INF;
    front = 0, rear = 1;
    hm.init();
    hm.push(0, -1);
    while(front != rear)
    {
        x = qx[front], y = qy[front];
        for(i = 0; i < 4; i ++)
        {
            newx = x + dx[i], newy = y + dy[i];
            if(newx >= 0 && newx < 8 && newy >= 0 && newy < 8)
            {
                qx[rear] = newx, qy[rear] = newy;
                for(j = 0; j < 6; j ++)
                    st[rear][ch[i][j]] = st[front][j];
                v[rear] = v[front] + st[rear][4];
                if(k = hm.push(rear, where[front]))
                {
                    where[rear] = k;
                    ++ rear;
                    if(rear > Q)
                        rear = 0;
                }
            }
        }
        ++ front;
        if(front > Q)
            front = 0;
    }
    for(i = 1; i < hm.e; i ++)
        if(hm.x[i] == tx && hm.y[i] == ty && hm.f[i] < ans)
            ans = hm.f[i], k = i;
    printf("%d", ans);
    print(k);
    printf("\n");
}
int main()
{
    int i;
    while(scanf("%s", b) == 1)
    {
        sx = b[0] - 'a', sy = b[1] - '1';
        scanf("%s", b);
        tx = b[0] - 'a', ty = b[1] - '1';
        qx[0] = sx, qy[0] = sy;
        for(i = 0; i < 6; i ++)
            scanf("%d", &st[0][i]);
        v[0] = st[0][4], where[0] = 1;
        solve();
    }
    return 0;
}