USACO Section 1.4 The Clocks（DFS）

The Clocks
IOI'94 - Day 2

Consider nine clocks arranged in a 3x3 array thusly:

|-------|    |-------|    |-------|    
|       |    |       |    |   |   |    
|---O   |    |---O   |    |   O   |          
|       |    |       |    |       |           
|-------|    |-------|    |-------|    
    A            B            C

|-------|    |-------|    |-------|
|       |    |       |    |       |
|   O   |    |   O   |    |   O   |
|   |   |    |   |   |    |   |   |
|-------|    |-------|    |-------|
    D            E            F

|-------|    |-------|    |-------|
|       |    |       |    |       |
|   O   |    |   O---|    |   O   |
|   |   |    |       |    |   |   |
|-------|    |-------|    |-------|
    G            H            I

The goal is to find a minimal sequence of moves to return all the dials to 12 o'clock. Nine different ways to turn the dials on the clocks are supplied via a table below; each way is called a move. Select for each move a number 1 through 9 which will cause the dials of the affected clocks (see next table) to be turned 90 degrees clockwise.

Move	Affected clocks
1	ABDE
2	ABC
3	BCEF
4	ADG
5	BDEFH
6	CFI
7	DEGH
8	GHI
9	EFHI

Example

Each number represents a time accoring to following table:

9 9 12       9 12 12       9 12 12        12 12 12      12 12 12 
6 6 6  5 ->  9  9  9  8->  9  9  9  4 ->  12  9  9  9-> 12 12 12 
6 3 6        6  6  6       9  9  9        12  9  9      12 12 12 

[But this might or might not be the `correct' answer; see below.]

PROGRAM NAME: clocks

INPUT FORMAT

Lines 1-3:

Three lines of three space-separated numbers; each number represents the start time of one clock, 3, 6, 9, or 12. The ordering of the numbers corresponds to the first example above.

SAMPLE INPUT (file clocks.in)

9 9 12
6 6 6
6 3 6

OUTPUT FORMAT

A single line that contains a space separated list of the shortest sequence of moves (designated by numbers) which returns all the clocks to 12:00. If there is more than one solution, print the one which gives the lowest number when the moves are concatenated (e.g., 5 2 4 6 < 9 3 1 1).

SAMPLE OUTPUT (file clocks.out)

4 5 8 9

/*
ID:xxx111_1
LANG:C
TASK:clocks
*/


#include<stdio.h>
#include<stdlib.h>
#include<string.h>


int  a[4][4],b[100],ans,num,c,flag,t;

void A(int p)
{
        a[1][1]=(a[1][1]+p)%4;
}

void B(int p)
{
        a[1][2]=(a[1][2]+p)%4;
}

void C(int p)
{
        a[1][3]=(a[1][3]+p)%4;
}

void D(int p)
{
        a[2][1]=(a[2][1]+p)%4;
}

void E(int p)
{
        a[2][2]=(a[2][2]+p)%4;
}

void F(int p)
{
        a[2][3]=(a[2][3]+p)%4;
}

void G(int p)
{
        a[3][1]=(a[3][1]+p)%4;
}

void H(int p)
{
        a[3][2]=(a[3][2]+p)%4;
}

void I(int p)
{
        a[3][3]=(a[3][3]+p)%4;
}

int step(int x,int p)
{
    int i;
    for  (i=1;  i<=p ;  i++)
    {
        num++;
        b[num]=x;
    }
    if  (x==1)
    {
        A(p); B(p); D(p); E(p);
    }
    if  (x==2)
    {
        A(p); B(p); C(p);
    }
    if  (x==3)
    {
        B(p); C(p); E(p); F(p);
    }
    if  (x==4)
    {
        A(p); D(p); G(p);
    }
    if  (x==5)
    {
        B(p); D(p); E(p); F(p); H(p);
    }
    if  (x==6)
    {
        C(p); F(p); I(p);
    }
    if  (x==7)
    {
        D(p); E(p); G(p); H(p);
    }
    if  (x==8)
    {
        G(p); H(p); I(p);
    }
    if  (x==9)
    {
        E(p); F(p); H(p); I(p);
    }
}


void check()
{
    int  i,j;
    for  (i=1; i<=3; i++)
       for (j=1; j<=3;  j++)
         if (a[i][j]!=0)    return;
    flag=1;           return;
}


void dfs(int x)
{
    int i,j,c;

    check();
    if  (flag==1 )
    {
        for (i=1; i<num; i++)
           printf("%d ",b[i]);
        printf("%d\n",b[num]);
        fclose(stdin);
        fclose(stdout);
        exit(0);
    }

    /*printf("!!!%d %d\n",x,flag);
    for (j=1; j<=3; j++)
    {
        printf("%d %d %d \n",a[j][1],a[j][2],a[j][3]);
    }
    printf("\n");

    scanf("%d",&c);*/
    if  ( (x==10) || (flag==1)  )  return;
    for  (i=0;  i<=3; i++)
    {
        int  tmp[4][4];
        memcpy(tmp,a,sizeof(a));
        step(x,i);
        dfs(x+1);
        memcpy(a,tmp,sizeof(tmp));
        num=num-i;
    }

}

int main()
{
   // freopen("clocks.in","r",stdin);
   // freopen("clocks.out","w",stdout);
    int  i;
    for  (i=1; i<=3; i++)
    {
       scanf("%d %d %d",&a[i][1],&a[i][2],&a[i][3]);
       a[i][1]=(a[i][1]/3)%4;
       a[i][2]=(a[i][2]/3)%4;
       a[i][3]=(a[i][3]/3)%4;
    }
    num=0;  flag=0;
    dfs(1);
    return 0;
}