poj2965The Pilots Brothers' refrigerator

背景：和poj1753一样，用dfs就可以做出来，只是和1753相比较得输出一些步骤，不过这也不麻烦，直接用两个数组就可以存储了。不过记住当递归回来的时候记住把数组里面对应位置的元素清零。

思路：同上一篇1753.

#include <stdio.h>
#include <string.h>
int q[4][4],ok=0,r[16],c[16];
int iswin(void)
{for(int i=0;i<4;i++)for(int j=0;j<4;j++)if(q[i][j]!=1) return 0;return 1;
}
void change(int i,int j)
{q[i][j]=!q[i][j];for(int k=0;k<4;k++){q[i][k]=!q[i][k];q[k][j]=!q[k][j];}
}
void dfs(int m,int sum,int a,int b)
{if(ok) return;else if(m==sum){if(iswin()){ok=1;printf("%d\n",sum);for(int k=0;k<sum;k++)printf("%d %d\n",r[k],c[k]);}else return;}else{for(int i=a;i<4;i++){for(int j=b;j<4;j++){r[m]=i+1;c[m]=j+1;change(i,j);if(j<3)  dfs(m+1,sum,i,j+1);else {dfs(m+1,sum,i+1,0);b=0;}change(i,j);r[m]=0;c[m]=0;}}return;}
}
int main(void)
{char ch;memset(r,0,sizeof(r));memset(c,0,sizeof(c));for(int i=0;i<4;i++)for(int j=0;j<4;j++){scanf("%c",&ch);if(j==3) getchar();q[i][j]=(ch=='+')?0:1;}if(iswin()) {printf("%d\n",0);return 0;}for(int k=1;k<=16;k++){dfs(0,k,0,0);if(ok) break;}return 0;
}

在网上看到有这样用递归的，感觉好高大上。

虽然他写的代码，我有一些地方不懂，不过这些都是我要学习的东西。

他的思路：

1.一个元素翻转奇数次状态不变,翻转偶数次状态改变

由1容易推到出2

2:要想把第Sij翻转,同时保持第i行和第j列其他元素状态不变,sij本身翻转两次,第i和第j列的其他元素翻转一次.

由把所有状态为关的元素都做一次2操作,记录那些元素状态变化过,变化过的元素个数即最小步骤数.

由以上分析得出,各个元素的步骤顺序其实是无所谓的.把变化的过元素按任意顺序输出即步骤.

附代码:

#include <iostream>
#include <queue>
bool mark[4][4];
struct Pos
{  int x;  int y;
};
std::queue<Pos> Ans;
int input()
{  memset(mark,0,sizeof(mark));  int i ,j,k;  char ch;  for (i = 0 ;i < 4 ; ++ i)  {  for (j = 0 ;j < 4 ; ++j)  {  ch = getchar();  if ('+' == ch)  {  mark[i][j] = !mark[i][j];  for(k = 0 ; k < 4 ; k ++)  {  mark[i][k] = !mark[i][k];  mark[k][j] = ! mark[k][j];  }  }  }  getchar();  }  return 0;
}
int main()
{  int nStep = 0;  int i,j;  input();  for ( i = 0 ; i < 4; ++i)  {  for (j = 0 ; j < 4 ; ++j)  {  if (mark[i][j])  {  nStep ++;  Pos pos;  pos.x = i + 1;  pos.y = j + 1;  Ans.push(pos);  }  }  }  std::cout <<nStep<<std::endl;  while (!Ans.empty())  {  Pos pos = Ans.front();  Ans.pop();  std::cout<<pos.x<<" "<<pos.y<<std::endl;  }  return 0;
}

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