1.4 clocks 枚举

注意转变思路，每个操作最多执行三次，所有解空间是4**9个，枚举所有可能的操作

钟表a受影响的操作有1，3，4

用上述方法，判断每个操作后的状态

输出即可

 

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

 

/* ID: xdzhbin1 LANG: C++ TASK: clocks */ //枚举 //每个操作执行的次数是0到3，枚举所有可能的操作 #include <fstream> #include <stack> #include <iostream> #include <vector> using namespace std; //把数字num转化为4进制数，表示每个操作的执行次数 void change(vector<int>& v,int num) { int i=8,j; while(num!=0) { v[i] = num % 4; num /= 4; i--; } for(j=0;j<=i;j++) { v[j]=0; } } //是否到达目标状态 bool check(int a[]) { for(int i=0; i<9; i++) { if(a[i]!=0&&a[i]!=12) return false; } return true; } int main() { ifstream in("clocks.in",ios::in); ofstream out("clocks.out",ios::out); int array[9],a[9]; int i,j; for(i=0; i<9; i++) { in>>array[i]; //读入初始各个钟表的状态 } vector<int> v(9,0); for(i=0; i<262144; i++) { change(v,i); for(j=0;j<9;j++) { a[j]=array[j]; } a[0]+=(v[0]+v[1]+v[3])*3%12; //时钟a受操作0，1，3的影响 a[1]+=(v[0]+v[1]+v[2]+v[4])*3%12; a[2]+=(v[1]+v[2]+v[5])*3%12; a[3]+=(v[0]+v[3]+v[4]+v[6])*3%12; a[4]+=(v[0]+v[2]+v[4]+v[6]+v[8])*3%12; a[5]+=(v[2]+v[4]+v[5]+v[8])*3%12; a[6]+=(v[3]+v[6]+v[7])*3%12; a[7]+=(v[4]+v[6]+v[7]+v[8])*3%12; a[8]+=(v[5]+v[7]+v[8])*3%12; if(check(a)) break; } bool space = false; //标识是否输出空格 for(i=0; i<9; i++) { for(j=0; j<v[i]; j++) { if(space) out<<" "; out<<i+1; space = true; } } out<<endl; return 0; }