poj算法-使用C语言在VC++6.0实现poj2965(枚举)

poj2965:

The Pilots Brothers’ refrigerator

题目描述：
he game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a refrigerator.

There are 16 handles on the refrigerator door. Every handle can be in one of two states: open or closed. The refrigerator is open only when all handles are open. The handles are represented as a matrix 4х4. You can change the state of a handle in any location [i, j] (1 ≤ i, j ≤ 4). However, this also changes states of all handles in row i and all handles in column j.

The task is to determine the minimum number of handle switching necessary to open the refrigerator.

输入：

The input contains four lines. Each of the four lines contains four characters describing the initial state of appropriate handles. A symbol “+” means that the handle is in closed state, whereas the symbol “−” means “open”. At least one of the handles is initially closed.

输出：

The first line of the input contains N – the minimum number of switching. The rest N lines describe switching sequence. Each of the lines contains a row number and a column number of the matrix separated by one or more spaces. If there are several solutions, you may give any one of them.

例如，输入：(引号内为实际输入)
“-+–”
“—-”
“—-”
“-+–”

输出：
6
1 1
1 3
1 4
4 1
4 3
4 4

一个冰箱上有4*4共16个开关，改变任意一个开关的状态（即开变成关，关变成开）时，此开关的同一行、同一列所有的开关都会自动改变状态。要想打开冰箱，要所有开关全部打开才行。
输入：一个4×4的矩阵，+表示关闭，-表示打开；
输出：使冰箱打开所需要执行的最少操作次数，以及所操作的开关坐标。

#include <stdio.h>
#include <string.h>

char handle[4][4];//存储字符
int record[4][4]; //记录每一位操作次数

//函数：输入，即设置开关
void set_handle()
{
    int i;
    for(i = 0; i < 4; i++)
    {
        scanf("%s", handle[i]);
    }
    memset(record, 0, sizeof(record));
}

char change_state(char state)
{
    if(state == '+')
        return '-';
    return '+';
}

//函数：把m行n列的开关按下
void flip(int m, int n)
{
    int i, j;
    for(j = 0; j < 4; j++) //行操作
    {
        handle[m][j] = change_state(handle[m][j]);
    }
    for(i = 0; i < 4; i++) //列操作
    {
        if(i != m)
            handle[i][n] = change_state(handle[i][n]);
    }
    record[m][n]++; //执行一次操作
}

//函数：对数组中所有为+位置的行和列所有的开关执行一次flip操作
void full_flip()
{
    int i, j;
    for(i = 0; i < 4; i++)
    {
        for(j = 0; j < 4; j++)
        {
            if(handle[i][j] == '+')
            {
                int t;
                for(t = 0; t < 4; t++)
                {
                    flip(i, t);
                }

                for(t = 0; t < 4; t++)
                {
                    if(t != i)
                    {
                        flip(t,j);
                    }

                }
            }
        }
    }
}


//函数：遍历record数组，统计操作次数
void solute()
{
    int count = 0;
    int i, j;
    for(i = 0; i < 4; i++)
    {
        for(j = 0; j < 4; j++)
        {
            if(record[i][j] & 1 != 0)
            {
                count ++;
            }
        }
    }
    printf("%d\n", count);
    for(i = 0; i < 4; i++)
    {
        for(j = 0; j < 4; j++)
        {
            if(record[i][j] & 1 != 0)
            {
                printf("%d %d\n", i+1, j+1);
            }
        }
    }

}

int main(void)
{
    set_handle();
    full_flip();
    solute();
    return 0;
}

代码在VC++6.0中正确运行，也可以在poj中正确运行。