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.

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

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

  1 /*
  2 LANG: C++
  3 PROG: clocks
  4 */
  5 #include <iostream>
  6 #include <cstdio>
  7 #include <algorithm>
  8 #include <cstring>
  9 #include <cmath>
 10 #define siji(i,x,y) for(int i=(x) ; i <= (y) ; ++i)
 11 #define xiaosiji(i,x,y) for(int i = (x) ; i < (y); ++i)
 12 #define gongzi(j,x,y) for(int j = (x) ; j >= (y) ; --j)
 13 #define ivorysi
 14 #define mo 11447
 15 #define eps 1e-8
 16 #define o(x) ((x)*(x))
 17 using namespace std;
 18 typedef long long ll;
 19 int clo[4][4],rec[10],ans[10],all=10000;
 20 void span(int &a) {
 21     a=(a+1)%4;
 22 }
 23 bool check() {
 24     siji(i,1,3) {
 25         siji(j,1,3) {
 26             if(clo[i][j]!=3)return false;
 27         }
 28     }
 29     return true;
 30 }
 31 void del1() {
 32     span(clo[1][1]);
 33     span(clo[1][2]);
 34     span(clo[2][1]);
 35     span(clo[2][2]);
 36 }
 37 void del2() {
 38     span(clo[1][1]);
 39     span(clo[1][2]);
 40     span(clo[1][3]);
 41 }
 42 void del3() {
 43     span(clo[1][2]);
 44     span(clo[1][3]);
 45     span(clo[2][2]);
 46     span(clo[2][3]);
 47 }
 48 void del4() {
 49     span(clo[1][1]);
 50     span(clo[2][1]);
 51     span(clo[3][1]);
 52 }
 53 void del5() {
 54     span(clo[1][2]);
 55     span(clo[2][1]);
 56     span(clo[2][2]);
 57     span(clo[2][3]);
 58     span(clo[3][2]);
 59 }
 60 void del6() {
 61     span(clo[1][3]);
 62     span(clo[2][3]);
 63     span(clo[3][3]);
 64 }
 65 void del7() {
 66     span(clo[2][1]);
 67     span(clo[2][2]);
 68     span(clo[3][1]);
 69     span(clo[3][2]);
 70 }
 71 void del8() {
 72     span(clo[3][1]);
 73     span(clo[3][2]);
 74     span(clo[3][3]);
 75 }
 76 void del9() {
 77     span(clo[2][2]);
 78     span(clo[2][3]);
 79     span(clo[3][2]);
 80     span(clo[3][3]);
 81 }
 82 void dfs(int dep,int times) {
 83     if(times>=all) return;
 84     if(dep>9) {
 85         if(!check()) return;
 86         all=times;
 87         memcpy(ans,rec,sizeof(rec));
 88         return;
 89     }
 90     void (*cur)();
 91     if(dep==1) cur=&del1;
 92     else if(dep==2) cur=&del2;
 93     else if(dep==3) cur=&del3;
 94     else if(dep==4) cur=&del4;
 95     else if(dep==5) cur=&del5;
 96     else if(dep==6) cur=&del6;
 97     else if(dep==7) cur=&del7;
 98     else if(dep==8) cur=&del8;
 99     else if(dep==9) cur=&del9;
100
101     siji(i,1,3) {
102         (*cur)();
103         rec[dep]=i;
104         dfs(dep+1,times+i);
105     }
106     (*cur)();
107     rec[dep]=0;
108     dfs(dep+1,times);
109 }
110 void solve() {
111     siji(i,1,3) {
112         siji(j,1,3) {
113             scanf("%d",&clo[i][j]);
114             clo[i][j]/=3;
115             --clo[i][j];
116         }
117     }
118     int cnt=0;
119     dfs(1,0);
120     siji(i,1,9) {
121         siji(j,1,ans[i]) {
122             ++cnt;
123             printf("%d%c",i," \n"[cnt==all]);
124         }
125     }
126 }
127 int main(int argc, char const *argv[])
128 {
129 #ifdef ivorysi
130     freopen("clocks.in","r",stdin);
131     freopen("clocks.out","w",stdout);
132 #else
133     freopen("f1.in","r",stdin);
134     //freopen("f1.out","w",stdout);
135 #endif
136     solve();
137     return 0;
138 }