fengshui

2010-06-09 01:06:51

Judge Info

• Memory Limit: 32768KB
• Case Time Limit: 10000MS
• 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  northeastwest                   eastsouthwest  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

2GOLD WOOD WATERWATER     FIREWOOD GOLD GROUNDWATER GOLD WOODWOOD       WATERGOLD GROUND GROUNDWATER GROUND WOODGOLD         FIREGOLD  FIRE   GROUNDGOLD  FIRE   FIREGOLD         FIREWATER GROUND WOOD

Sample Output

2 14

用BFS，但可能会超时，所以用了双向BFS

==参考代码==

 

#include <stdio.h>#include <memory.h>int fig[8],pos[400000],BFS[400000],flag[400000];int fff[8]={1,5,25,125,625,3125,15625,78125}; int f(char str[]){if(str[2]=='L')return 0;if(str[2]=='T')return 3;if(str[2]=='R')return 4;if(str[3]=='D')return 1;return 2;} void input(){char a[7],b[7],c[7];scanf("%s%s%s",a,b,c);fig[0]=f(a);fig[1]=f(b);fig[2]=f(c);scanf("%s%s",a,b);fig[3]=f(b);fig[7]=f(a);scanf("%s%s%s",a,b,c);fig[6]=f(a);fig[5]=f(b);fig[4]=f(c);} int encode(){int i,sum;sum=0;for(i=0;i<8;i++)sum+=fig[i]*fff[i];return sum;} void discode(int sum){int i;for(i=0;i<8;i++){fig[i]=sum%5;sum/=5;}} int solve(int st,int ed){int top,base,i,j,now,next,max0,max1;base=0;top=0;if(st==ed)return 0;BFS[top++]=st;flag[st]=0;pos[st]=0;max0=1;BFS[top++]=ed;flag[ed]=1;pos[ed]=1;max1=1;while(top>base){now=BFS[base++];discode(now);if(flag[now]==0){max0--;next=(now%fff[7])*5+now/fff[7];if(flag[next]==1)return pos[now]+pos[next];if(pos[next]==-1){pos[next]=pos[now]+1;flag[next]=0;BFS[top++]=next;max0++;}for(i=0;i<8;i++)for(j=0;j<8;j++)if((fig[j]+1)%5==fig[i]){next=now-(fig[i]-(fig[i]+3)%5)*fff[i];if(flag[next]==1)return pos[now]+pos[next];if(pos[next]==-1){pos[next]=pos[now]+1;flag[next]=0;BFS[top++]=next;max0++;}break;}if(max0<=0)return -1;} else {max1--;next=(now%5)*fff[7]+now/5;if(flag[next]==0)return pos[now]+pos[next];if(pos[next]==-1){pos[next]=pos[now]+1;flag[next]=1;BFS[top++]=next;max1++;}for(i=0;i<8;i++)for(j=0;j<8;j++)if(fig[j]==(fig[i]+1)%5){next=now-(fig[i]-(fig[i]+2)%5)*fff[i];if(flag[next]==0)return pos[now]+pos[next];if(pos[next]==-1){pos[next]=pos[now]+1;flag[next]=1;BFS[top++]=next;max1++;}break;}if(max1<=0)return -1;} }return -1;} void init(){memset(pos,-1,sizeof(pos));memset(flag,-1,sizeof(flag));} int main(){int i,test,st,ed;scanf("%d",&test);while(test--){input();st=encode();input();ed=encode();init();printf("%d/n",solve(st,ed));}return 0;}