poj2965--The Pilots Brothers' refrigerator

最新推荐文章于 2021-10-24 15:01:26 发布

原创最新推荐文章于 2021-10-24 15:01:26 发布 · 634 阅读

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本文介绍了一个基于矩阵操作的游戏谜题解决方案。玩家需要通过切换冰箱门上16个把手的状态来打开冰箱，每次操作会改变同一行和列的把手状态。文章提供了一段C语言代码，演示了如何找到最小的操作步骤。

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The Pilots Brothers' refrigerator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17642		Accepted: 6685		Special Judge

Description

The 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.

Input

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.

Output

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.

Sample Input

-+--
----
----
-+--

Sample Output

Source

Northeastern Europe 2004, Western Subregion

和翻转棋差不多的做法，只是改为了翻转一行和一列

#include <stdio.h>
#include <string.h>
struct node{
    int a[16] ;
    int t , n ;
    int s[16] ;
    int top ;
}p[1000000];
int main()
{
    int i , j , a[16] , min_t = 999999 , to ;
    char str[4][4] ;
    for(i = 0 ; i < 4 ; i++)
        scanf("%s", str[i]);
    for(i = 0 ; i < 4 ; i++)
        for(j = 0 ; j < 4 ; j++)
            if(str[i][j] == '+')
                a[4*i+j] = 1 ;
            else
                a[4*i+j] = 0 ;
    int low = 0 , top = 1 ;
    for(i = 0 ; i < 16 ; i++)
        p[0].a[i] = a[i] ;
    p[0].t = 0 ; ;
    p[0].n = -1 ;
    p[0].top = 0 ;
    while(low < top)
    {
        node k = p[low++] ;
        int num = 0 ;
        for(i = 0 ; i < 16 ; i++)
            num += k.a[i] ;
        if(num == 0 && k.t < min_t)
        {
            min_t = k.t ;
            to = low-1 ;
        }
        if(k.n == 15)
            continue ;
        if(k.t >= min_t)
            continue ;
        p[top] = k ;
        p[top].n++ ;
        top++ ;
        p[top] = k ;
        p[top].n++ ;
        p[top].t++ ;
        p[top].s[ p[top].top++ ] = p[top].n ;
        i = p[top].n / 4 ;
        j = p[top].n % 4 ;
        int s ;
        for(s = 0 ; s < 4 ; s++)
        {
            p[top].a[4*i+s] = 1 - p[top].a[4*i+s] ;
            p[top].a[4*s+j] = 1 - p[top].a[4*s+j] ;
        }
        p[top].a[4*i+j] = 1 - p[top].a[4*i+j] ;
        top++ ;
    }
    printf("%d\n", p[to].top);
    int q ;
    for(q = 0 ; q < p[to].top ; q++)
    {
        i = p[to].s[q] / 4 ;
        j = p[to].s[q] % 4 ;
        i++;
        j++ ;
        printf("%d %d\n", i , j );
    }
    return 0;
}