• POJ 1753 Flip Game 暴力 深搜


    Flip Game
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 59468   Accepted: 24750

    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
    

    Source

     
     
      1 #include <iostream>
      2 #include <cstdio>
      3 
      4 using namespace std;
      5 
      6 char str[6];
      7 int a[6][6];
      8 int ans=17;
      9 
     10 void flip(int x,int y)
     11 {
     12     a[x][y]=!a[x][y];
     13     a[x-1][y]=!a[x-1][y];
     14     a[x+1][y]=!a[x+1][y];
     15     a[x][y-1]=!a[x][y-1];
     16     a[x][y+1]=!a[x][y+1];
     17 }
     18 
     19 bool same_color(void)
     20 {
     21     for(int i=1;i<=4;++i)
     22     {
     23         for(int j=1;j<=4;++j)
     24         {
     25             if(a[i][j]!=a[1][1])
     26             {
     27                 return false;
     28             }
     29         }
     30     }
     31     return true;
     32 }
     33 
     34 void dfs(int x,int y,int t)
     35 {
     36     // 判断是否同色,同色满足,返回
     37     if(same_color())
     38     {
     39         if(ans>t)
     40         {
     41             ans=t;
     42         }
     43         return;
     44     }
     45     // 深搜截止条件
     46     if(x>4)
     47     {
     48         return;
     49     }
     50 
     51     if(y==4)
     52     {
     53         // 当前不反转
     54         dfs(x+1,1,t);
     55         // 当前反转
     56         flip(x,y);
     57         dfs(x+1,1,t+1);
     58         // 状态还原
     59         flip(x,y);
     60     }
     61     else
     62     {
     63         dfs(x,y+1,t);
     64         flip(x,y);
     65         dfs(x,y+1,t+1);
     66         flip(x,y);
     67     }
     68 
     69 }
     70 
     71 int main()
     72 {
     73     for(int i=1;i<=4;++i)
     74     {
     75         scanf("%s",str+1);
     76         for(int j=1;j<=4;++j)
     77         {
     78             // a->0,b->1
     79             if(str[j]=='w')
     80             {
     81                 a[i][j]=0;
     82             }
     83             else
     84             {
     85                 a[i][j]=1;
     86             }
     87             //a[i][j]=str[j]-'a';
     88             //printf("%d ",a[i][j]);
     89         }
     90         //printf("
    ");
     91     }
     92 
     93 
     94     dfs(1,1,0);
     95 
     96     if(ans==17)
     97     {
     98         printf("Impossible
    ");
     99     }
    100     else
    101     {
    102         printf("%d
    ",ans);
    103     }
    104 
    105 
    106 
    107     return 0;
    108 }
  • 相关阅读:
    Jmail组件发送邮件说明ASP.NET
    五种常见的ASP.NET应用程序安全缺陷
    按比例生成缩略图
    光盘自动运行HTML页,Autorun文件写法
    页面全屏显示JS代码
    除去内容中的HTML代码方法
    JS函数学习(2)
    JS学习变量与基本语法(1)
    C#中Math.Round()实现中国式四舍五入
    (2) EFCore数据库上下文和模型(数据表)的配置
  • 原文地址:https://www.cnblogs.com/jishuren/p/12370371.html
Copyright © 2020-2023  润新知