ACM POJ 2965 The Pilots Brothers' refrigerator

http://poj.org/problem?id=2965

The Pilots Brothers' refrigerator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10158		Accepted: 3707		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

6
1 1
1 3
1 4
4 1
4 3
4 4

Source

Northeastern Europe 2004, Western Subregion

 

 

题目分析：电冰箱有16个把手，每个把手两种状态（开‘－’或关‘＋’），只有在所有把手都打开时，门才开，输入数据是个4＊4的矩阵，因此考虑用位表示。可以改变任意一个把手的位置，但同时改变其所在的行和列。求最小步骤，因此算法是位图＋广搜。如果当前状态在队列中已出现，则无需再次入队，因此需要判重。一个2^16-1个状态，在0时结束退出。需要有一个变量记录当前的状态，以及一个变量由队列中那个位置转移过来。在一个状态后共有16个状态改变方案。

位运算，BFS，然后注意记录路径！

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
const int MAXN=65536;
int que[MAXN*2];
int pre[MAXN];
int prei[MAXN];
int prej[MAXN];
int step[MAXN];

void bfs(int p)
{
    int i,j,rear, front;
   
    rear=front=0;
     que[rear++]=p;
    memset(pre,-1,sizeof(pre));
    memset(step,0,sizeof(step));
    memset(prei,-1,sizeof(prei));
    memset(prej,-1,sizeof(prej));
    pre[p]=p;
    //step[p]++;
    while(front<rear)
    {
        int tmp=que[front++];
        int t=tmp;
       for(i=0;i<=3;i++)
          for(j=0;j<=3;j++) 
          {
              tmp=t;
              tmp^=1<<(15-j);
              tmp^=1<<(15-j-4);
              tmp^=1<<(15-j-8);
              tmp^=1<<(15-j-12);
              tmp^=1<<(15-4*i);
              tmp^=1<<(15-4*i-1);
              tmp^=1<<(15-4*i-2);
              tmp^=1<<(15-4*i-3);
              
              tmp^=1<<(15-4*i-j);//(i,j)多倒了一次，倒回去
              
               
              if(pre[tmp]==-1)
              {
                  pre[tmp]=t;
                  step[tmp]=step[t]+1;
                  prei[tmp]=i+1;
                  prej[tmp]=j+1;
                  que[rear++]=tmp;
              }     
              if(tmp==0)
              {
                  printf("%d\n",step[tmp]);
                  int x,y,n;
                  int a[MAXN],b[MAXN];
                  n=x=y=step[tmp];
                  while(pre[tmp]!=tmp)
                  {
                      a[x--]=prei[tmp];
                      b[y--]=prej[tmp];
                      tmp=pre[tmp];
                  } 
                  for(int k=1;k<=n;k++)
                  {
                      printf("%d %d\n",a[k],b[k]);
                  }  
                  return;     
              }    
          }    
    }    
    
}   
int main()
{
    freopen("test.in","r",stdin);
    freopen("test.out","w",stdout);
    int n=0;
    char ch;
    int i,j;
    for(i=0;i<4;i++)
      for(j=0;j<4;j++)
      {
          cin>>ch;n<<=1;
          if(ch=='+')n+=1;
          
      }    
    bfs(n);
    return 0;    
    
}