HDU 1667 The Rotation Game

HDU_1667

    如果我们确定要移动一类整数的话，那么其他两类整数实际上都可以看作是0，因此可以固定一个中间八个整数为1、其余整数为0的终态，然后预处理出所有可能的状态，在查询的时候只要枚举1、2、3分别作为要移动到中间整数就可以计算出最优解了。

#include<stdio.h>
#include<string.h>
#define HASH 1000007
#define MAXD 1000010
#define INF 0x3f3f3f3f
int op[][24] =
{
    {22, 1, 0, 3, 4, 5, 2, 7, 8, 9, 10, 6, 12, 13, 14, 11, 16, 17, 18, 19, 15, 21, 20, 23},
    {0, 23, 2, 1, 4, 5, 6, 7, 3, 9, 10, 11, 8, 13, 14, 15, 16, 12, 18, 19, 20, 17, 22, 21},
    {0, 1, 2, 3, 5, 6, 7, 8, 9, 10, 4, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23},
    {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 19, 13, 20, 21, 22, 23},
    {0, 3, 2, 8, 4, 5, 6, 7, 12, 9, 10, 11, 17, 13, 14, 15, 16, 21, 18, 19, 20, 23, 22, 1},
    {2, 1, 6, 3, 4, 5, 11, 7, 8, 9, 10, 15, 12, 13, 14, 20, 16, 17, 18, 19, 22, 21, 0, 23},
    {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 19, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23},
    {0, 1, 2, 3, 10, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23},
};
int ini[] = {0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0};
struct HashMap
{
    int head[HASH], size, next[MAXD], st[MAXD];
    void init()
    {
        memset(head, -1, sizeof(head)), size = 0;
    }
    int find(int _st)
    {
        int i, h = _st % HASH;
        for(i = head[h]; i != -1; i = next[i])
            if(st[i] == _st) break;
        return i;
    }
    void push(int _st)
    {
        int h = _st % HASH;
        st[size] = _st;
        next[size] = head[h], head[h] = size ++;
    }
}hm;
int pre[MAXD], type[MAXD], dis[MAXD], q[MAXD];
int code[25], g[25];
char list[MAXD], b[MAXD];
int encode(int *code)
{
    int i, ans = 0;
    for(i = 0; i < 24; i ++) ans = ans << 1 | code[i];
    return ans;
}
void decode(int st)
{
    int i;
    for(i = 23; i >= 0; i --) code[i] = st & 1, st >>= 1;
}
void prepare()
{
    int i, j, k, x, y, rear = 0, t[25];
    hm.init();
    x = encode(ini);
    pre[0] = -1, dis[0] = 0;
    hm.push(x), q[rear ++] = x;
    for(i = 0; i < rear; i ++)
    {
        x = q[i];
        decode(x);
        for(j = 0; j < 8; j ++)
        {
            for(k = 0; k < 24; k ++) t[k] = code[op[j][k]];
            y = encode(t);
            k = hm.find(y);
            if(k == -1)
            {
                pre[rear] = i, dis[rear] = dis[i] + 1, type[rear]= j;
                hm.push(y), q[rear ++] = y;
            }
            else if(dis[i] + 1 == dis[k] && j < type[k])
                pre[k] = i, type[k] = j;
        }
    }
}
void print(int cur, char *q, int rear)
{
    if(pre[cur] == -1) return;
    q[rear] = type[cur] + 'A';
    print(pre[cur], q, rear + 1);
}
void solve()
{
    int i, j, t[25], ans = INF, cur, mid;
    for(i = 1; i < 24; i ++) scanf("%d", &g[i]);
    for(i = 1; i <= 3; i ++)
    {
        for(j = 0; j < 24; j ++) t[j] = g[j] == i;
        j = hm.find(encode(t));
        if(j != -1)
        {
            if(dis[j] < ans)
                ans = dis[j], cur = j, mid = i, print(cur, list, 0), list[ans] = '\0';
            else if(dis[j] == ans)
            {
                print(j, b, 0), b[ans] = '\0';
                if(strcmp(b, list) == -1)
                    strcpy(list, b), cur = j, mid = i;
            }
        }
    }
    if(ans == 0) printf("No moves needed\n%d\n", mid);
    else printf("%s\n%d\n", list, mid);
}
int main()
{
    int t;
    prepare();
    while(scanf("%d", &t), t)
    {
        g[0] = t;
        solve();
    }
    return 0;
}