USACO1.4.2--The Clocks

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
解题思路：DFS。每种方案最多选三次，因为旋转四次就恢复原状态了，相当于没进行操作。由于方案的顺序并不重要，因此满足要求就可以输出，中断搜索。
表示被一个BUG坑死了，整整坑了两天。仅仅因为把赋值符号“=”打成“==”了，然后就一直不输出结果，惨痛的教训啊，以后一定要记住！！！

View Code

#include <stdio.h>

int a[9][9]= { {3,3,3,3,3,2,3,2,0},
{2,3,2,3,2,3,1,0,1},
{3,3,3,2,3,3,0,2,3},
{2,3,1,3,2,0,2,3,1},
{2,3,2,3,1,3,2,3,2},
{1,3,2,0,2,3,1,3,2},
{3,2,0,3,3,2,3,3,3},
{1,0,1,3,2,3,2,3,2},
{0,2,3,2,3,3,3,3,3} };
int v[9];

int main() {
int i,j,k;
freopen("clocks.in","r",stdin);
for (i=0; i<9; i++) {
scanf("%d",&k);
for(j=0; j<9; j++)
v[j]=(v[j]+(4-k/3)*a[i][j])%4;
}
fclose(stdin);

k=0;
freopen("clocks.out","w",stdout);
for (i=0; i<9; i++)
for (j=0; j<v[i]; j++)
if (!k) { printf("%d",i+1); k=1; }
else    printf(" %d",i+1);
printf("\n");
fclose(stdout);
return 0;
}