操作间没有次序关系,同一个操作最多重复3次。。。

可以直接暴力。。。

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 according 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

Submission file Name:

USACO Gateway |   
Comment or Question

/*
ID:qhn9992
PROG:clocks
LANG:C++11
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm> using namespace std; int a[10],c[10];
int b[10][10]
={ 0,0,0,0,0,0,0,0,0,0,
1,1,0,1,1,0,0,0,0,0,
1,1,1,0,0,0,0,0,0,0,
0,1,1,0,1,1,0,0,0,0,
1,0,0,1,0,0,1,0,0,0,
0,1,0,1,1,1,0,1,0,0,
0,0,1,0,0,1,0,0,1,0,
0,0,0,1,1,0,1,1,0,0,
0,0,0,0,0,0,1,1,1,0,
0,0,0,0,1,1,0,1,1,0
}; int main()
{
freopen("clocks.in","r",stdin);
freopen("clocks.out","w",stdout);
int cnt=0; for(int i=1;i<=9;i++)
{
scanf("%d",a+i);
a[i]=(a[i]/3)%4;
}
for(c[1]=0;c[1]<4;c[1]++)
{
for(c[2]=0;c[2]<4;c[2]++)
{
for(c[3]=0;c[3]<4;c[3]++)
{
for(c[4]=0;c[4]<4;c[4]++)
{
for(c[5]=0;c[5]<4;c[5]++)
{
for(c[6]=0;c[6]<4;c[6]++)
{
for(c[7]=0;c[7]<4;c[7]++)
{
for(c[8]=0;c[8]<4;c[8]++)
{
for(c[9]=0;c[9]<4;c[9]++)
{
///check....
bool flag=true;
for(int i=1;i<=9;i++)
{
int t=a[i];
for(int j=1;j<=9;j++)
{
t+=b[j][i-1]*c[j];
}
t=t%4;
if(t)
{
flag=false;
break;
}
}
if(flag)
{
for(int i=1;i<=9;i++)
{
while(c[i]--)
{
if(flag)
{
flag=false;
}
else putchar(32);
printf("%d",i);
}
}
putchar(10);
return 0;
}
else continue;
}
}
}
}
}
}
}
}
}
return 0;
}

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