POJ 1166 The Clocks（暴搜）-优快云博客

本文探讨解决钟面所有指针同时指向12点的最短操作序列问题，通过遍历所有可能的组合来求解。使用了暴力搜索算法进行求解，并讨论了在特定条件下的优化可能性。

The Clocks

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12403		Accepted: 4908

Description

|-------|    |-------|    |-------|
|       |    |       |    |   |   |
|---O   |    |---O   |    |   O   |
|       |    |       |    |       |
|-------|    |-------|    |-------|
    A            B            C    

|-------|    |-------|    |-------|
|       |    |       |    |       |
|   O   |    |   O   |    |   O   |
|   |   |    |   |   |    |   |   |
|-------|    |-------|    |-------|
    D            E            F    

|-------|    |-------|    |-------|
|       |    |       |    |       |
|   O   |    |   O---|    |   O   |
|   |   |    |       |    |   |   |
|-------|    |-------|    |-------|
    G            H            I    
(Figure 1)

There are nine clocks in a 3*3 array (figure 1). The goal is to return all the dials to 12 o'clock with as few moves as possible. There are nine different allowed ways to turn the dials on the clocks. Each such way is called a move. Select for each move a number 1 to 9. That number will turn the dials 90' (degrees) clockwise on those clocks which are affected according to figure 2 below.

Move   Affected clocks

 1         ABDE
 2         ABC
 3         BCEF
 4         ADG
 5         BDEFH
 6         CFI
 7         DEGH
 8         GHI
 9         EFHI    
   (Figure 2)

Input

Your program is to read from standard input. Nine numbers give the start positions of the dials. 0=12 o'clock, 1=3 o'clock, 2=6 o'clock, 3=9 o'clock.

Output

Your program is to write to standard output. Output a shortest sorted sequence of moves (numbers), which returns all the dials to 12 o'clock. You are convinced that the answer is unique.

Sample Input

3 3 0
2 2 2
2 1 2

Sample Output

4 5 8 9

Source

IOI 1994

这题没意思。。。暴搜就过了。。

本来想练习高斯消元的，却发现4不是素数会出问题。。

好像可以用逆矩阵做的。。。还在思考中

#include <string.h>
#include <stdio.h>
int main()
{
      int i,a[10],b[10],c[10];
      for(i=1;i<=9;i++)
      scanf("%d",&a[i]);
      for(b[1]=0;b[1]<=3;b[1]++)
      for(b[2]=0;b[2]<=3;b[2]++)
      for(b[3]=0;b[3]<=3;b[3]++)
      for(b[4]=0;b[4]<=3;b[4]++)
      for(b[5]=0;b[5]<=3;b[5]++)
      for(b[6]=0;b[6]<=3;b[6]++)
      for(b[7]=0;b[7]<=3;b[7]++)
      for(b[8]=0;b[8]<=3;b[8]++)
      for(b[9]=0;b[9]<=3;b[9]++)  
      {
            c[1]=(a[1]+b[1]+b[2]+b[4])%4;
            c[2]=(a[2]+b[1]+b[2]+b[3]+b[5])%4;
            c[3]=(a[3]+b[2]+b[3]+b[6])%4;
            c[4]=(a[4]+b[1]+b[4]+b[5]+b[7])%4;
            c[5]=(a[5]+b[1]+b[3]+b[5]+b[7]+b[9])%4;
            c[6]=(a[6]+b[3]+b[5]+b[6]+b[9])%4;
            c[7]=(a[7]+b[4]+b[7]+b[8])%4;
            c[8]=(a[8]+b[5]+b[7]+b[8]+b[9])%4;
            c[9]=(a[9]+b[6]+b[8]+b[9])%4;
            if(c[1]+c[2]+c[3]+c[4]+c[5]+c[6]+c[7]+c[8]+c[9]==0)
            {
                    for(i=0;i<b[1];i++) printf("1 ");
                    for(i=0;i<b[2];i++) printf("2 ");
                    for(i=0;i<b[3];i++) printf("3 ");
                    for(i=0;i<b[4];i++) printf("4 ");
                    for(i=0;i<b[5];i++) printf("5 ");
                    for(i=0;i<b[6];i++) printf("6 ");
                    for(i=0;i<b[7];i++) printf("7 ");
                    for(i=0;i<b[8];i++) printf("8 ");
                    for(i=0;i<b[9];i++) printf("9 ");
                    printf("\n");
                    return(0);
            }
      }
}