标签：c++
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
题意：类似 1753 每次可以翻棋但同行同列都要翻 求最小步数 并输出翻棋达到全部是"—"的顺(不一定)序；
代码：
#include<iostream>
#include<cstdio>
using namespace std;
int step1[18][3];
int step2[18][3];
bool map2[5][5];
char map1[5][5];
int s;
int min1;
bool judge()
{
int i,j;
for(i=0; i<4; i++)
for(j=0; j<4; j++)
if(map2[i][j]==false)
return false;
return true;
}
void dfs(int x,int y)
{
int i,j;
for(i=0;i<4;i++)
for(j=0;j<4;j++)
{
if(i==x)
map2[i][j]=!map2[i][j];
if(j==y)
map2[i][j]=!map2[i][j];
}
map2[x][y]=!map2[x][y];
}
void work(int k,int step)
{
int i;
if(k==16)
{
if(judge()&&step<min1)
{
min1=step;
for(i=0;i<min1;i++)
step2[i][0]=step1[i][0],step2[i][1]=step1[i][1]; //关键 保存步数
}
}
else
{
int x=k/4;
int y=k%4;
work(k+1,step);
dfs(x,y);
step1[step][0]=x;
step1[step][1]=y;
work(k+1,step+1);
dfs(x,y);
}
}
int main()
{
int i,j;
for(i=0; i<4; i++)
for(j=0; j<4; j++)
{
cin>>map1[i][j];
if(map1[i][j]=='+')
map2[i][j]=false;
else
map2[i][j]=true;
}
min1=17;
work(0,0);
cout<<min1<<endl;
for(i=0;i<min1;i++)
cout<<step2[i][0]+1<<' '<<step2[i][1]+1<<endl;
return 0;
}
POJ 2965 The Pilots Brothers' refrigerator 搜索+枚举,布布扣,bubuko.com
POJ 2965 The Pilots Brothers' refrigerator 搜索+枚举
标签：c++
原文地址：http://blog.csdn.net/axuan_k/article/details/37914321
