POJ1166 The Clocks （爆搜 || 高斯消元）

总时间限制: 1000ms，内存限制: 65536kB

描述

|-------|    |-------|    |-------|
|       |    |       |    |   |   |
|---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)

输入

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.

输出

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.

样例输入

3 3 0
2 2 2
2 1 2

样例输出

4 5 8 9

学枚举时候的课程作业，第一次接触到爆搜的魅力，实在是香。不过这题还可以用高斯消元，也是学到了。

爆搜代码

#include <string.h>
#include <stdio.h>
int main()
{
    int i, a[10], b[10], c[10]; //多开一位空间，安全，b[i]就是第i种Move，c[i]就是Move之后第i个时钟的状态 
    for (i = 1; i <= 9; i++)
        scanf("%d", &a[i]); //读入初始状态 
    //枚举每一种移动方法，每一种可实施0,1,2,3次
    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);
                                        }
                                    }
}

高斯消元做法链接：

https://blog.csdn.net/sf____/article/details/9863927