IOI‘94 - Day 2
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A B C
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D E F
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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
思路：9种操作中，没有一种是对单独一个钟进行旋转的，那么能否找到一种操作组合，这一操作组合使一个钟顺时针旋转90度？
设我们对操作i使用了a[i]次（0<=a[i]<=3，因为当a[i]=4时的效果和a[i]=0时一样），那么要只让时钟A顺时针旋转90度的话，由表知，有(a[1]+a[2]+a[4])%4=1,(a[1]+a[2]+a[3]+a[5])%4=0,...使用如下暴力代码可以得到全部的操作次数：
#include<cstdio>
#include<cstring>
#define For(x) for (a[x] = 0; a[x] < 4; ++a[x])
const char C[9][9] =
{
"ABDE",
"ABC",
"BCEF",
"ADG",
"BDEFH",
"CFI",
"DEGH",
"GHI",
"EFHI"
};
bool clocks[9][9];
int a[9], cnt[9];
int main()
{
puts("const int A[9][9] =\n{");
int i, j, choose;
for (i = 0; i < 9; ++i)
for (j = 0; C[i][j]; ++j)
clocks[i][C[i][j] - ‘A‘] = true;
for (choose = 0; choose < 9; ++choose)
{
For(0)For(1)For(2)For(3)For(4)For(5)For(6)For(7)For(8)
{
memset(cnt, 0, sizeof(cnt));
for (i = 0; i < 9; ++i)
for (j = 0; j < 9; ++j)
if (clocks[j][i]) cnt[i] += a[j];
for (i = 0; i < 9; ++i)
{
if (i == choose)
{
if ((cnt[i] & 3) != 1) break;///x&3=x%4
}
else
{
if (cnt[i] & 3) break;
}
}
if (i == 9)
{
putchar(‘{‘);
for (i = 0; i < 8; ++i) printf("%d,", a[i]);
printf("%d}%s", a[8], (choose == 8 ? "\n" : ",\n"));
goto L;
}
}
L:
;
}
puts("};");
return 0;
}
随后使用之：
/*
ID: synapse cheng
PROG: clocks
LANG: C++
*/
/*
Executing...
Test 1: TEST OK [0.000 secs, 3376 KB]
Test 2: TEST OK [0.000 secs, 3376 KB]
Test 3: TEST OK [0.000 secs, 3376 KB]
Test 4: TEST OK [0.000 secs, 3376 KB]
Test 5: TEST OK [0.000 secs, 3376 KB]
Test 6: TEST OK [0.000 secs, 3376 KB]
Test 7: TEST OK [0.000 secs, 3376 KB]
Test 8: TEST OK [0.000 secs, 3376 KB]
Test 9: TEST OK [0.000 secs, 3376 KB]
All tests OK.
*/
#include<cstdio>
const 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 main()
{
freopen("clocks.in", "r", stdin);
freopen("clocks.out", "w", stdout);
int v[9] = {0}, i, j, k;
for (i = 0; i < 9; i++)
{
scanf("%d", &k);
for (j = 0; j < 9; j++) v[j] += (4 - k / 3) * A[i][j];
}
for (i = 0; i < 9; i++) v[i] &= 3;
k = 0;
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);
putchar(10);
return 0;
}