1. 这道题有好多种解法，看NOCOW，用枚举，DFS，BFS都能求解。我看到这个题的时候，连最基本的枚举怎么做都想不出来，差距呐。。。；
2. 我挑选了解题报告中代码最短的读懂然后写出来了：
/*
ID: dollar4
PROG: clocks
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <string>
#include <algorithm>
#include <cstring>
#include <numeric>
#include <climits>
using namespace std;
#define FOR(x) for (c[x] = 0; c[x] <= 3; c[x]++)
int main()
{
ofstream fout ("clocks.out");
ifstream fin ("clocks.in");
int i, c[10], a[10], temp, s[10], rst[10], cnt, mincnt = INT_MAX;
for (i = 0; i < 9; i++)
{
fin >> temp;
a[i] = temp / 3 % 4;
}
FOR(1) FOR(2) FOR(3) FOR(4) FOR(5) FOR(6) FOR(7) FOR(8) FOR(9)
{
s[0] = (a[0] + c[1] + c[2] + c[4]) % 4;
s[1] = (a[1] + c[1] + c[2] + c[3] + c[5]) % 4;
s[2] = (a[2] + c[2] + c[3] + c[6]) % 4;
s[3] = (a[3] + c[1] + c[4] + c[5] + c[7]) % 4;
s[4] = (a[4] + c[1] + c[3] + c[5] + c[7] + c[9]) % 4;
s[5] = (a[5] + c[3] + c[5] + c[6] + c[9]) % 4;
s[6] = (a[6] + c[4] + c[7] + c[8]) % 4;
s[7] = (a[7] + c[5] + c[7] + c[8] + c[9]) % 4;
s[8] = (a[8] + c[6] + c[8] + c[9]) % 4;
if (accumulate(s, s + 9, 0))
continue;
cnt = accumulate(c + 1, c + 10, 0);
if (cnt < mincnt)
{
mincnt = cnt;
copy(c + 1, c + 10, rst);
}
}
for (i = 0; i < 9; i++)
{
while (rst[i]--)
{
fout << i + 1;
if (--cnt)
fout << ' ';
else fout << endl;
}
}
return 0;
}
3. 官方参考代码：
Notice that the order in which we apply moves is irrelevant, and that applying a move four times is the same as applying it not at all.
Thus there are only 49 = 262144 move sequences that need to be tried, so we might as well just try them all.
We don't generate them shortest first, but looking at sequences of the same length, we generate the lesser ones before the greater ones, so we only need to keep track of the shortest working sequence we've found.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <ctype.h>
#define INF 60000 /* bigger than longest possible path */
char *movestr[] = { "abde", "abc", "bcef", "adg", "bdefh", "cfi", "degh",
"ghi", "efhi" };
int movedist[9][9];
int clock[9];
int bestmove[9];
int nbestmove;
/* translate move strings into array "movedist" of which clocks change on each move */
void
mkmove(void)
{
int i;
char *p;
for(i=0; i<9; i++)
for(p=movestr[i]; *p; p++)
movedist[i][*p-'a'] = 3;
}
/* apply some number of each move from k to 9 */
/* move contains the number of times each move is applied */
void
solve(int *move, int k)
{
int i, j, n, rep;
if(k == 9) {
for(j=0; j<9; j++)
if(clock[j]%12 != 0)
return;
/* we have a successful sequence of moves */
n = 0;
for(j=0; j<9; j++)
n += move[j];
if(nbestmove == 0 || n < nbestmove) {
nbestmove = n;
for(j=0; j<9; j++)
bestmove[j] = move[j];
}
return;
}
/*
* the for loop counts down so we
* generate smaller numbers first by
* trying more of small numbered
* moves before we try less of them.
*/
for(rep=3; rep>=0; rep--) {
/* apply move k rep times */
for(i=0; i<rep; i++)
for(j=0; j<9; j++)
clock[j] += movedist[k][j];
move[k] = rep;
solve(move, k+1);
/* undo move */
for(i=0; i<rep; i++)
for(j=0; j<9; j++)
clock[j] -= movedist[k][j];
}
}
void
main(void)
{
FILE *fin, *fout;
int i, j, move[9];
char *sep;
fin = fopen("clocks.in", "r");
fout = fopen("clocks.out", "w");
assert(fin != NULL && fout != NULL);
mkmove();
for(i=0; i<9; i++)
fscanf(fin, "%d", &clock[i]);
solve(move, 0);
sep = "";
for(i=0; i<9; i++) {
for(j=0; j<bestmove[i]; j++) {
fprintf(fout, "%s%d", sep, i+1);
sep = " ";
}
}
fprintf(fout, "\n");
exit(0);
}Lucian Boca submitted a constant time solution
You can precalculate a matrix a as following:
a[i][0],a[i][1],....,a[i][8] is the number of moves '1','2','3',...'9' necessarly to move ONLY clock i (where 0 < i <= 8 - there are 9 clocks: 0=A, 1=B, ... 8=I) 90 degrees clockwise. So, you have the matrix:
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} };
That means: to move ONLY the clock 0 (clock A) 90 degrees clockwise you have to do {3,3,3,3,3,2,3,2,0}, 3 moves of type 1, three moves of type 2, ..., 2 moves of type 8, 0 moves of type 9, etc.
To move ONLY the clock 8 (clock I), you have to do the moves {0,2,3,2,3,3,3,3,3}: 0 moves of type 1, 2 moves of type 2... 3 moves of type 9....
That's it! You count in a vector v[9] how many moves of each type you have to do, and the results will be modulo 4 (%4 - 5 moves of any type have the same effect 1 move has). The source 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;
}Here's a different approach from Sopot Cela -- no loops, constant time. It is, however, extraordinarily intricate: getting such a program correct in a contest environment could be extremely challenging:
program clocks;
const
INV : array[3..12] of byte =(1, 0, 0, 2, 0, 0, 3, 0, 0, 0);
var inp, out: text;
a, b, c, d, e, f, g, h, i: integer;
ax, bx, cx, dx, ex, fx, gx, hx, ix,l: integer;
t,an: array[1..9] of integer;
begin
assign (inp, 'clocks.in'); reset (inp);
readln(inp, ax, bx, cx);
readln(inp, dx, ex, fx);
readln(inp, gx, hx, ix);
a:=inv[ax]; b:=inv[bx]; c:=inv[cx]; d:=inv[dx];
e:=inv[ex]; f:=inv[fx]; g:=inv[gx]; h:=inv[hx];
i:=inv[ix];
t[1] := (8+a+2*b+c+2*d+2*e-f+g-h) mod 4;
t[2] := (a+b+c+d+e+f+2*g+ 2*i) mod 4;
t[3] := (8+ a+2*b+ c -d+2*e+2*f -h+ i) mod 4;
t[4] := ( a+ b+2*c+ d+ e+ g+ h+2*i) mod 4;
t[5] := (4+ a+2*b+ c+2*d -e+2*f+ g+2*h+ i) mod 4;
t[6] := ( 2*a+ b+ c+ e+ f+2*g+ h+ i) mod 4;
t[7] := (8+ a -b+ 2*d+2*e -f+ g+2*h+ i) mod 4;
t[8] := ( 2*a+ 2*c+ d+ e+ f+ g+ h+ i) mod 4;
t[9] := (8 -b+ c -d+2*e+2*f+ g+2*h+ i) mod 4;
assign(out, 'clocks.out'); rewrite(out);
for a := 1 to 9 do
for b := 1 to t[a] do Begin
inc(l);
an[l]:=a;
end;
for a:=1 to l-1 do
write(out,an[a],' ');
write(out,an[l]);
writeln(out); close(out)
end.