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

题意：通过上面九种操作把给定的矩阵变成全是12。
分析：首先要理解最终转变成 12 和操作的顺序无关，然后就是每个操作最多只能做 3 次，因为第 4 次又回到初始状态。所以依次从操作 1 开始枚举到操作 9，同时每层要枚举操作的次数依次是3 2 1 0。最后就是输出了要注意一下，最后不能有空格。

 /*
 ID: dizzy_l1
 LANG: C++
 TASK: clocks
 */
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>

using namespace std;

int op[10][5]={{0,0,0,0,0},
               {1,2,4,5,0},{1,2,3,0,0},{2,3,5,6,0},
               {1,4,7,0,0},{2,4,5,6,8},{3,6,9,0,0},
               {4,5,7,8,0},{7,8,9,0,0},{5,6,8,9,0},
              };
int a[10],ans[10];
bool flag;

bool judge()
{
    int i;
    for(i=1;i<10;i++)
        if(a[i])
            return false;
    return true;
}

void output(int p)//最后不能有空格
{
    int i,j,k;
    k=p;
    while(ans[k]==0) k--;
    for(i=1;i<k;i++)
    {
        for(j=0;j<ans[i];j++)
            printf("%d ",i);
    }
    for(j=0;j<ans[k]-1;j++)
        printf("%d ",k);
    printf("%d\n",k);
    flag=true;
}

void dfs(int p)
{
    if(judge())
    {
        output(p);return ;
    }
    if(flag||p>9) return ;
    int i,j;
    for(i=3;i>=0;i--)
    {
        int ta[10];
        ans[p]=i;
        memcpy(ta,a,sizeof(a));
        for(j=0;j<5;j++)
            if(op[p][j])
                a[op[p][j]]=(a[op[p][j]]+i)%4;
        dfs(p+1);
        memcpy(a,ta,sizeof(ta));
    }
}

int main()
{
    //freopen("clocks.in","r",stdin);
    //freopen("clocks.out","w",stdout);
    int i;
    for(i=1;i<10;i++)
    {
        scanf("%d",&a[i]);
        a[i]=(a[i]/3)%4;
    }
    flag=false;
    dfs(1);
    return 0;
}