• 枚举(黑白棋)


    /*代码一:DFS+Enum*/
    
    //Memory Time 
    //240K   344MS 
    
    //本题只要求输出翻转的次数,因此BFS或DFS都适用
    
    #include<iostream>
    using namespace std;
    
    bool chess[6][6]={false};//利用的只有中心的4x4
    bool flag;
    int step;
    int r[]={-1,1,0,0,0};//便于翻棋操作
    int c[]={0,0,-1,1,0};
    
    bool judge_all(void)//判断“清一色”
    {
        int i,j;
        for(i=1;i<5;i++)
            for(j=1;j<5;j++)
                if(chess[i][j]!=chess[1][1])
                    return false;
        return true;
    }
    
    void flip(int row,int col)//翻棋
    {
        int i;
        for(i=0;i<5;i++)
            chess[row+r[i]][col+c[i]]=!chess[row+r[i]][col+c[i]];
        return;
    }
    
    void dfs(int row,int col,int deep) //深搜的迭代回溯是重点,很容易混乱
    {
        if(deep==step)
        {
            flag=judge_all();
            return;
        }
    
        if(flag||row==5)return;
    
        flip(row,col);       //翻棋
        if(col<4)
            dfs(row,col+1,deep+1);
        else
            dfs(row+1,1,deep+1);
    
        flip(row,col);      //不符合则翻回来
        if(col<4)
            dfs(row,col+1,deep);
        else
            dfs(row+1,1,deep);
    
        return;
    }
    
    int main(void)
    {
        char temp;
        int i,j;
        for(i=1;i<5;i++)
            for(j=1;j<5;j++)
            {
                cin>>temp;
                if(temp=='b') 
                    chess[i][j]=true;
            }
    
        for(step=0;step<=16;step++)  //对每一步产生的可能性进行枚举
        {                            //至于为什么是16,考虑到4x4=16格,而每一格只有黑白两种情况,则全部的可能性为2^16
            dfs(1,1,0);
            if(flag)break;
        }
    
        if(flag)
            cout<<step<<endl;
        else
            cout<<"Impossible"<<endl;
        return 0;
    }
    View Code
    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

  • 相关阅读:
    临时生成的文件下载“提速” (提早开始下载)(node.js)
    node.js http接口调试时请求串行特性分析
    presto 判断数据量是否大于一个比较小的值的优化
    typescript 关于class属性类型定义被属性默认值覆盖的问题及解决方式
    typescript 属性默认值使用箭头函数 this指向问题
    presto 函数中使用子查询
    浅谈redis的HyperLogLog与布隆过滤器
    关于管理后台更新与响应的设计
    对“算法-求二进制数中1的个数” 中一些要点的补充
    关于async 中return 和 return await 的差异
  • 原文地址:https://www.cnblogs.com/baoluqi/p/3745308.html
Copyright © 2020-2023  润新知