zoukankan      html  css  js  c++  java
  • poj 1753 Flip Game

    http://poj.org/problem?id=1753

    题意:4*4的黑白棋子,给定棋子现在的状态怎样翻棋使棋子变成全黑or全白,但是每翻一次棋子,它的四周的都会改变成相反的棋子,是黑就会变白,是白就会变黑。

    分析:一共最多有16步,所以可以枚举这16步,在每一种几步进行搜索,比如0步去搜索,1步去搜索可以不,2步去搜索可以不。。。。。

    枚举+dfs

                                                                                          Flip Game
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 31763   Accepted: 13822

    Description

    Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the other one is black and each piece is lying either it's black or white side up. Each round you flip 3 to 5 pieces, thus changing the color of their upper side from black to white and vice versa. The pieces to be flipped are chosen every round according to the following rules: 
    1. Choose any one of the 16 pieces. 
    2. Flip the chosen piece and also all adjacent pieces to the left, to the right, to the top, and to the bottom of the chosen piece (if there are any).
    Consider the following position as an example: 
    bwbw  wwww  bbwb  bwwb  Here "b" denotes pieces lying their black side up and "w" denotes pieces lying their white side up. If we choose to flip the 1st piece from the 3rd row (this choice is shown at the picture), then the field will become: 
    bwbw  bwww  wwwb  wwwb  The goal of the game is to flip either all pieces white side up or all pieces black side up. You are to write a program that will search for the minimum number of rounds needed to achieve this goal. 

    Input

    The input consists of 4 lines with 4 characters "w" or "b" each that denote game field position.

    Output

    Write to the output file a single integer number - the minimum number of rounds needed to achieve the goal of the game from the given position. If the goal is initially achieved, then write 0. If it's impossible to achieve the goal, then write the word "Impossible" (without quotes).

    Sample Input

    bwwb
    bbwb
    bwwb
    bwww

    Sample Output

    4
     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 using namespace std;
     5 int ans[6][6]={0};
     6 int r[]={1,0,-1,0};
     7 int c[]={0,1,0,-1};
     8 int step;
     9 bool flag=false;
    10 bool judge()
    11 {
    12     int i,j,sum=0;
    13      for(i=1;i<=4;i++)
    14         for(j=1;j<=4;j++)
    15             sum+=ans[i][j];
    16       if(sum==0||sum==16)
    17            return true;
    18        else
    19            return false;
    20 }
    21 void flip(int x,int y)
    22 {
    23        int i;
    24       if(ans[x][y]==1)
    25            ans[x][y]=0;
    26       else
    27          ans[x][y]=1;
    28     for(i=0;i<=3;i++)
    29      {
    30       if(ans[x+r[i]][y+c[i]]==0)
    31             ans[x+r[i]][y+c[i]]=1;
    32        else
    33          ans[x+r[i]][y+c[i]]=0;
    34      }
    35 }
    36 void dfs(int x,int y,int d)
    37 {
    38      if(d==step)
    39      {
    40        flag=judge();
    41            return ;
    42      }
    43 
    44      if(flag||x>4)
    45        {
    46         return ;
    47        }
    48     flip(x,y);//翻棋
    49      if(y<4)
    50       dfs(x,y+1,d+1);
    51      else
    52        dfs(x+1,1,d+1);
    53      flip(x,y);//翻回来.
    54      if(y<4)
    55       dfs(x,y+1,d);
    56      else
    57        dfs(x+1,1,d);
    58      
    59 }
    60 int main()
    61 {
    62 
    63        int i,j;
    64       char temp;
    65       for(i=1;i<=4;i++)
    66       {
    67         for(j=1;j<=4;j++)
    68              {
    69                  scanf("%c",&temp);
    70                  if(temp=='b')
    71                       ans[i][j]=1;
    72              }
    73              getchar();
    74        }
    75      for(step=0;step<=16;step++)//枚举16种步。
    76        {
    77 
    78           dfs(1,1,0);
    79            if(flag)
    80               break;
    81         }
    82       if(flag)
    83          printf("%d
    ",step);
    84       else
    85          printf("Impossible
    ");
    86      return 0;
    87 }
  • 相关阅读:
    Stage划分原理
    Spark转换和动作算子
    Spark运行原理
    Scrapy数据持久化
    在实际工作中Eclipse常用的快捷键
    关于Linux(CentOS6.5)中文输入法、中文桌面可视化等问题
    Eclipse无法启动报An internal error occurred during: "reload maven project". java.lang.NullPointerExceptio错原因及解析
    在Eclipse中复制导入项目并且修改原来的项目名字,项目后面的括号显示原来项目的名字!
    response.sendRedirect()与request.getRequestDispatcher("/index.jsp").forward(request, response)两者辨析
    MyEclipse/Eclipse导出jar方法
  • 原文地址:https://www.cnblogs.com/cancangood/p/4127305.html
Copyright © 2011-2022 走看看