POJ 2965 The Pilots Brothers' refrigerator （枚举）

2019-11-08 03:11:15

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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

61 11 31 44 14 34 4

题意

把题目中给出的状态图,全部翻转成 - 的状态

每次翻转一个，它所在的行和列都要进行翻转

问最小翻转次数，同时输出翻转路径。

思路

和 POJ 1753 一样，只不过在dfs的时候用一个栈或者双端队列存储当前的路径，找到答案之后输出。

AC 代码

#include <iostream>#include<stdio.h>#include<math.h>#include<string.h>#include<algorithm>#include<stdlib.h>#include<queue>using namespace std;bool bits[16];bool flag=false;deque<int>sk;void rese(int n) //翻转{ int x=n/4,y=n%4; for(int i=0; i<4; i++) bits[x*4+i]=!bits[x*4+i]; for(int i=0; i<4; i++) if(i!=x) bits[i*4+y]=!bits[i*4+y];}bool jud() //判断是否一致{ bool ini=bits[0]; if(!ini)return false; for(int i=1; i<16; i++) if(ini!=bits[i]) return false; return true;}void solve(int maxx,int now,int step){ if(maxx==step) //翻转完最后一枚棋子 { if(jud()) //满足状态 { flag=true; printf("%d/n",maxx); for(int i=0; i<maxx; i++) { printf("%d %d/n",sk.front()/4+1,sk.front()%4+1); sk.pop_front(); } } return; } for(int i=now; i<16; i++) //从上次翻转位置继续 { rese(i); //翻转i sk.push_back(i); solve(maxx,i+1,step+1); if(flag)return; //找到答案，返回 sk.pop_back(); rese(i); //恢复原来状态 }}int main(){ char str[4][4]; for(int i=0; i<4; i++) { cin>>str[i]; for(int j=0; j<4; j++) bits[i*4+j]=(str[i][j]=='-')?1:0; } for(int i=0; i<=16; i++) //i 为要翻转的棋子个数 solve(i,0,0); return 0;}

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