Poj 2965 The Pilots Brothers‘ refrigerator

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 openhe the refrigerator.

在poj 1753的基础上增加了路径保存，递归输出即可。

#include<iostream>
#include<limits.h>
#include<string.h>
using namespace std;
int sta=0;
bool mi[65536];
short int st[65536];
int ori[65536];
bool is_end(int status,int step){if (status==0) {cout<<step<<endl;return(true);}else return(false);
}
int work2(int n,int status){int tt=1<<n;return(status^tt);
}
int work(int n,int status){int temp=status;int ii=n/4,jj=n%4;for (int i=0;i<4;i++)temp=work2(i*4+jj,temp);for (int i=0;i<4;i++)temp=work2(ii*4+i,temp);temp=work2(n,temp);return(temp);
}
void bfs(int status){int head=0,tail=0,temp,step,temp_status;int que1[65536],que2[65536];bool t1=true;que1[0]=status;que2[0]=0;while (head<=tail){temp=que1[head];step=que2[head];head++;for (int i=15;i>=0;i--){temp_status=work(i,temp);       if (!mi[temp_status]){mi[temp_status]=true;st[temp_status]=i;ori[temp_status]=temp;tail++;que1[tail]=temp_status;que2[tail]=step+1;}if (is_end(temp_status,step+1)) {t1=false;break;}}if (!t1) break; }if (t1) cout<<"Impossible"<<endl;
}
void print(int status){if (status!=sta){print(ori[status]);cout<<(4-st[status]/4)<<' '<<(4-st[status]%4)<<endl;}
}
int main(){char ch;memset(mi,sizeof(mi),false);for(int i=0;i<16;i++){cin>>ch;sta=sta<<1;sta+=(ch=='+');}mi[sta]=true;bfs(sta);print(0);return 0;
}

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