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
This problem seems like a kind of for beginners.The time limit is up to 30 seconds.Or maybe it wasn't in the competition where it from.So beginners like me can pass this question through the most simple search.
The second operate reborn means for a element in some direction,if there exist a element which is also one of the eight directions and can kill it ,it can execute the reborn operate.
/*Gold 0
Wood 1
Water 2
Fire 3
Ground 4*/
/*
012
7 3
654
01234567
701
6 2
543
70123456
*/
#include<stdio.h>
#include<string.h>
#include<queue>
#include<iostream>
using namespace std;
struct node
{
int num,step;
};
const int pow5[10]={1,5,25,125,625,3125,15625,78125,390625,1953125};
const int kill[5] ={1,4,3,0,2};
const int born[5] ={2,3,1,4,0};
bool f[500000];
int map[8],S,T;
int bfs()
{
int i,j,k;
memset(f,0,sizeof(f));
f[S]=true;
queue<node> q;
while (!q.empty()) q.pop();
node tmp;
tmp.num=S;
tmp.step=0;
q.push(tmp);
while (!q.empty())
{
node x=q.front();
q.pop();
if (x.num==T) return x.step;
tmp=x;
tmp.step++;
int fr=tmp.num%5;
tmp.num=(tmp.num/5)+fr*pow5[7];
if (!f[tmp.num])
{
f[tmp.num]=true;
q.push(tmp);
}
tmp.num=x.num;
for (i=7;i>=0;i--)
{
map[i]=tmp.num%5;
tmp.num/=5;
}
for (i=0;i<8;i++)
{
bool flag=false;
for (j=0;j<8;j++)
if (kill[map[j]]==map[i])
{
int tt=map[i];
map[i]=born[map[i]];
tmp.num=0;
for (k=0;k<8;k++)
tmp.num=tmp.num*5+map[k];
if (!f[tmp.num])
{
f[tmp.num]=true;
q.push(tmp);
}
map[i]=tt;
break;
}
}
}
return -1;
}
int main()
{
int C,i;
scanf("%d",&C);
char s[8][15],t[8][15];
while (C--)
{
scanf("%s%s%s%s%s%s%s%s",s[0],s[1],s[2],s[7],s[3],s[6],s[5],s[4]);
scanf("%s%s%s%s%s%s%s%s",t[0],t[1],t[2],t[7],t[3],t[6],t[5],t[4]);
S=0,T=0;
for (i=0;i<8;i++)
{
if (s[i][0]=='G' && s[i][1]=='O') S=S*5+0;
if (s[i][0]=='W' && s[i][1]=='O') S=S*5+1;
if (s[i][0]=='W' && s[i][1]=='A') S=S*5+2;
if (s[i][0]=='F' && s[i][1]=='I') S=S*5+3;
if (s[i][0]=='G' && s[i][1]=='R') S=S*5+4;
}
for (i=0;i<8;i++)
{
if (t[i][0]=='G' && t[i][1]=='O') T=T*5+0;
if (t[i][0]=='W' && t[i][1]=='O') T=T*5+1;
if (t[i][0]=='W' && t[i][1]=='A') T=T*5+2;
if (t[i][0]=='F' && t[i][1]=='I') T=T*5+3;
if (t[i][0]=='G' && t[i][1]=='R') T=T*5+4;
}
printf("%d
",bfs());
}
return 0;
}