PKU 1753枚举

PKU 1753枚举

这个跟暑假做的一个题目是类似的 不过就是多了一种可能 全黑或者全白都可以

算法思路：
1 第一行枚举操作 下面的每行如果上行此列为黑（或者白）的时候进行操作
2 检查最后一行是否满足要求

剪枝：
如果操作次数已经超过当前最小次数 不必再试验

代码如下：

/**********************
Author: WHU_Victordu
Created Time: 2007-12-26
File Name: pku1753.cpp
  Description: 
   **************************/
#include "stdio.h"
#include "memory.h"
#define  INF 32675
int grid[4][4],a[4][4],out[4][4],op[4][4];
char cgrid[4][4];

void next()
{
      int j;
     for(j=3;j>=0;j--)
       {
           if(j==3)    op[0][j]++;
           else
           if(op[0][j+1]==2)    {op[0][j+1]=0;op[0][j]++;}
           else    break;
     }
}    

void copy()
{        
     int i,j;        
     for(i=0;i<4;i++)
         for(j=0;j<4;j++)
                 a[i][j]=grid[i][j];
 }

void change(int i,int j)
{
       a[i][j]^=1;
      if(j>0)   a[i][j-1]^=1;
      if(j<3) a[i][j+1]^=1;
      if(i>0)   a[i-1][j]^=1;
      if(i<3) a[i+1][j]^=1;
}    
          
int main()
{
      int i,j,cnt,min=INF,flag;
     
      for(i=0;i<4;i++)
      {
       gets(cgrid[i]);
      for(j=0;j<4;j++)
      {
      if(cgrid[i][j]=='w') grid[i][j]=0;
      else grid[i][j]=1;
      }
      }
      memset(op,0,sizeof(op));
      while(op[0][0]!=2)
      {
            copy();
            cnt=0;
            flag=0;
            for(i=0;i<=4;i++)
            {
                 if(i!=0&&i!=4)
                 {
                           for(j=0;j<4;j++)
                          if(a[i-1][j]==1)            
                          {
                                op[i][j]=1;
                                change(i,j);
                                cnt++;
                                if(cnt>min)    {flag=1;break;}
                           }
                           else    op[i][j]=0;
                  }                 
                  else
                  {
                           if(i==0)     
                         {
                           for(j=0;j<4;j++)    
                             if(op[i][j]==1)        
                           {
                                    change(i,j);cnt++;
                          }
                        }
                          else
                          if(i==4)
                         {
                              for(j=0;j<4&&a[i-1][j]==0;j++);    
                              if(j==4&&cnt<min)    {min=cnt;copy();}
                          }
                    }
                    if(flag==1)    break;
               }
                 next();
        }
       
      memset(op,0,sizeof(op));
      while(op[0][0]!=2)
      {
            copy();
            cnt=0;
            flag=0;
            for(i=0;i<=4;i++)
            {
                 if(i!=0&&i!=4)
                 {
                           for(j=0;j<4;j++)
                          if(a[i-1][j]==0)            
                          {
                                op[i][j]=1;
                                change(i,j);
                                cnt++;
                                if(cnt>min)    {flag=1;break;}
                           }
                           else    op[i][j]=0;
                  }                 
                  else
                  {
                           if(i==0)     
                         {
                           for(j=0;j<4;j++)    
                             if(op[i][j]==1)        
                           {
                                    change(i,j);cnt++;
                          }
                        }
                          else
                          if(i==4)
                         {
                              for(j=0;j<4&&a[i-1][j]==1;j++);    
                              if(j==4&&cnt<min)    {min=cnt;copy();}
                          }
                    }
                    if(flag==1)    break;
               }
                 next();
        }
           if(min==INF)    printf("Impossible\n");
           else
           printf("%d\n",min);
        // while(1);
         return 0;
}