USACO 6.5 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 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

————————————————————————————————————题解
4^9果断暴搜
然后秒过……
用了一个指向函数的指针减少代码量

 /*
 LANG: C++
 PROG: clocks
 */
 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 #define siji(i,x,y) for(int i=(x) ; i <= (y) ; ++i)
 #define xiaosiji(i,x,y) for(int i = (x) ; i < (y); ++i)
 #define gongzi(j,x,y) for(int j = (x) ; j >= (y) ; --j)
 #define ivorysi
 #define mo 11447
 #define eps 1e-8
 #define o(x) ((x)*(x))
 using namespace std;
 typedef long long ll;
 int clo[][],rec[],ans[],all=;
 void span(int &a) {
     a=(a+)%;
 }
 bool check() {
     siji(i,,) {
         siji(j,,) {
             if(clo[i][j]!=)return false;
         }
     }
     return true;
 }
 void del1() {
     span(clo[][]);
     span(clo[][]);
     span(clo[][]);
     span(clo[][]);
 }
 void del2() {
     span(clo[][]);
     span(clo[][]);
     span(clo[][]);
 }
 void del3() {
     span(clo[][]);
     span(clo[][]);
     span(clo[][]);
     span(clo[][]);
 }
 void del4() {
     span(clo[][]);
     span(clo[][]);
     span(clo[][]);
 }
 void del5() {
     span(clo[][]);
     span(clo[][]);
     span(clo[][]);
     span(clo[][]);
     span(clo[][]);
 }
 void del6() {
     span(clo[][]);
     span(clo[][]);
     span(clo[][]);
 }
 void del7() {
     span(clo[][]);
     span(clo[][]);
     span(clo[][]);
     span(clo[][]);
 }
 void del8() {
     span(clo[][]);
     span(clo[][]);
     span(clo[][]);
 }
 void del9() {
     span(clo[][]);
     span(clo[][]);
     span(clo[][]);
     span(clo[][]);
 }
 void dfs(int dep,int times) {
     if(times>=all) return;
     if(dep>) {
         if(!check()) return;
         all=times;
         memcpy(ans,rec,sizeof(rec));
         return;
     }
     void (*cur)();
     if(dep==) cur=&del1;
     else if(dep==) cur=&del2;
     else if(dep==) cur=&del3;
     else if(dep==) cur=&del4;
     else if(dep==) cur=&del5;
     else if(dep==) cur=&del6;
     else if(dep==) cur=&del7;
     else if(dep==) cur=&del8;
     else if(dep==) cur=&del9;

     siji(i,,) {
         (*cur)();
         rec[dep]=i;
         dfs(dep+,times+i);
     }
     (*cur)();
     rec[dep]=;
     dfs(dep+,times);
 }
 void solve() {
     siji(i,,) {
         siji(j,,) {
             scanf("%d",&clo[i][j]);
             clo[i][j]/=;
             --clo[i][j];
         }
     }
     int cnt=;
     dfs(,);
     siji(i,,) {
         siji(j,,ans[i]) {
             ++cnt;
             printf("%d%c",i," \n"[cnt==all]);
         }
     }
 }
 int main(int argc, char const *argv[])
 {
 #ifdef ivorysi
     freopen("clocks.in","r",stdin);
     freopen("clocks.out","w",stdout);
 #else
     freopen("f1.in","r",stdin);
     //freopen("f1.out","w",stdout);
 #endif
     solve();
     return ;
 }