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SZU Problem(B40):Fengshui

这道题超级坑，就是一个大大的坑，后台有三组数据，总题目的时间不能超过3秒，但是每组数组不能超过1秒，否则case time limited. 

因为我的方法很简单，就是套用BFS，八个方向为一个状态。。。不知道为什么，三组数据的总时间不超3秒，但她老人家第一数据就老超一秒，然后我就一直一直一直过不了。。

后来优化到已经不能再优化了，我就突然发现，我的queue的数组好像开得有点点大，这样每轮初始化花的时间也会增加，我就只给了他几乎刚刚的量，然后就AC啦~~

啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊阿啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊超开心^^

Judge Info

• Memory Limit: 32768KB
• Case Time Limit: 15000MS
• Time Limit: 30000MS
• Judger: Normal

Description

Fengshui is an ancient subject in Chinese tradition. Someone considers it as science and someone criticizes it as blind faith. Who knows! However, in modern days, everyone should respect culture from our ancestor!

Fengshui focus on geography,environment and staffs' position, all the theory come from a very old book named "YI". YI means change. Everything is always changing in the world. Fengshui wishes to guide changing, make life change to a better situation. Now let's look at Fengshui's changing.

At first we must know about the traditional five elements system which composed by GOLD,WOOD,GROUND,WATER and FIRE. Everything in the world can be represented by one and only one element. For example, river is represented by WATER, hill is represented by GROUND. Here, we only consider the elements. In this system, once element can kill another element, and one element can born anther element. Five elements compose as a circuit, as in Figure 1.

Every place has eight direction - east, west, north, south, northeast, northwest, southeast and southwest. Every direction has a represented element. Now, our problem is about the elements at these eight directions which form a Fengshui situation. Figure 2 is an example of one Fengshui situation.

• But Fengshui situation can change! There're two change ways:
  • TURN: The whole situation turn clockwise one step. Figure 3 shows the situation that situation in Figure 2 makes one TURN change.
  • REBORN: Based on kill and born relation, one direction's element can be killed by another direction's (at any other place) element in the situation, and then the killed element will born out as the new element at its direction. Of course, kill and born are all according as the relation of the system as in Figure 1. In situation of Figure 3, WATER in east can kill FIRE in southeast, then southeast place change to be GROUND, as in Figure 4.

Each change, no matter TURN or REBORN, const one step.

Now, there're two Fengshui situation, we want to know is it possible that first one can change to the second one? And if possible, how many steps it need at least?

Input

There're several cases, the first line of input is the number of cases. Every case includes 6 lines, the first 3 lines indeicate the first Fengshui situation, the last 3 lines incicate the second Fengshui situation.

The format of one situation is as follow, there may be arbitrary blanks between adjacent directions.

northwest  north  northeast
west                   east
southwest  south  southeast

Output

For every case, output the number of the least changing steps on a single line, if it is possible, or output -1.

Sample Input

2
GOLD WOOD WATER
WATER     FIRE
WOOD GOLD GROUND
WATER GOLD WOOD
WOOD       WATER
GOLD GROUND GROUND
WATER GROUND WOOD
GOLD         FIRE
GOLD  FIRE   GROUND
GOLD  FIRE   FIRE
GOLD         FIRE
WATER GROUND WOOD

Sample Output

2 
14

虽然代码很长很罗嗦，而且跑得很慢，刚刚卡三秒的时间，但我还是很开心啊哈哈哈哈哈哈~

#include <stdio.h>
#include <string.h>
#define L 5
#define N 8
#define M 3125050
int former[N];
int goal[N];
int sign[L];
int kill[] = { 3, 0, 4, 2, 1 };
int born[] = { 2, 3, 1, 4, 0 };
int distance[L][L][L][L][L][L][L][L];
int flag[L][L][L][L][L][L][L][L];
int queue[M];
int sequence[] = { 0, 1, 2, 7, 3, 6, 5, 4 };
 
int main()
{
    char temp[N];
    int t, i, j, k, p, head;
    scanf( "%d", &t );
    while( t-- )
    {
        // 初始化
        memset( distance, 0, sizeof(distance) );
        memset( flag, 0, sizeof(flag) );
        p = 0;
        for( i=0; i<N; ++i )
        {
            scanf( "%s", temp );
            if( temp[0] == 'G' && temp[1] == 'O' )
                former[sequence[p++]] = 0;
            else if( temp[0] == 'W' && temp[1] == 'O' )
                former[sequence[p++]] = 1;
            else if( temp[0] == 'W' && temp[1] == 'A' )
                former[sequence[p++]] = 2;
            else if( temp[0] == 'F' && temp[1] == 'I' )
                former[sequence[p++]] = 3;
            else
                former[sequence[p++]] = 4;
        }
        p = 0;
        for( i=0; i<N; ++i )
        {
            scanf( "%s", temp );
            if( temp[0] == 'G' && temp[1] == 'O' )
                goal[sequence[p++]] = 0;
            else if( temp[0] == 'W' && temp[1] == 'O' )
                goal[sequence[p++]] = 1;
            else if( temp[0] == 'W' && temp[1] == 'A' )
                goal[sequence[p++]] = 2;
            else if( temp[0] == 'F' && temp[1] == 'I' )
                goal[sequence[p++]] = 3;
            else
                goal[sequence[p++]] = 4;
        }
        head = p = 0;
        for( i=0; i<N; ++i )
            queue[p++] = former[i];
        flag[former[0]][former[1]][former[2]][former[3]][former[4]][former[5]][former[6]][former[7]] = 1;
        while( head < p )
        {
            memset( sign, 0, sizeof(sign) );
            for( i=0; i<N; ++i )
            {
                former[i] = queue[head+i];
                sign[former[i]] = 1;
            }
 
            int x = distance[former[0]][former[1]][former[2]][former[3]][former[4]][former[5]][former[6]][former[7]];
 
 
            // 顺时针旋转
            int temp = former[N-1];
            for( i=N-1; i>0; --i )
                former[i] = former[i-1];
 
            former[0] = temp;
 
            if( flag[former[0]][former[1]][former[2]][former[3]][former[4]][former[5]][former[6]][former[7]] == 0 )
            {
                distance[former[0]][former[1]][former[2]][former[3]][former[4]][former[5]][former[6]][former[7]] = x+1;
                flag[former[0]][former[1]][former[2]][former[3]][former[4]][former[5]][former[6]][former[7]] = 1;
 
                for( i=0; i<N; ++i )
                    if( former[i] != goal[i] )
                        break;
                if( i == N  ) // 一致跳出while循环
                    break;
                else // 不一致放入队列中去
                    for( i=0; i<N; i++ )
                        queue[p++] = former[i];
            }
            // 相克替换
 
            for( k=0; k<N; k++ ) // for一遍
            {
                if( sign[kill[queue[head+k]]] == 1 ) // 如果相克
                {
                            // 从队列中得到当前状态
                    for( j=0; j<N; ++j )
                        former[j] = queue[head+j];
                    former[k] = born[queue[head+k]]; // 进行替换
                    if( flag[former[0]][former[1]][former[2]][former[3]][former[4]][former[5]][former[6]][former[7]] == 0 )
                    {
                        distance[former[0]][former[1]][former[2]][former[3]][former[4]][former[5]][former[6]][former[7]] = x+1;
                        flag[former[0]][former[1]][former[2]][former[3]][former[4]][former[5]][former[6]][former[7]] = 1;
 
                        for( j=0; j<N; ++j )
                            if( former[j] != goal[j] )
                                break;
                        if( j == N )
                            goto END;
                        else
                            for( j=0; j<N; ++j )
                                queue[p++] = former[j];
                    }
                }
            }
            head += 8;
        }
END:    if( head < p )
            printf( "%d\n", distance[former[0]][former[1]][former[2]][former[3]][former[4]][former[5]][former[6]][former[7]] );
        else
            printf( "-1\n" );
    }
    return 0;
}