zoukankan      html  css  js  c++  java
  • POJ 2965 The Pilots Brothers' refrigerator 位运算枚举

     

     
    The Pilots Brothers' refrigerator
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 15176   Accepted: 5672   Special Judge

    Description

    The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a refrigerator.

    There are 16 handles on the refrigerator door. Every handle can be in one of two states: open or closed. The refrigerator is open only when all handles are open. The handles are represented as a matrix 4х4. You can change the state of a handle in any location [i, j] (1 ≤ i, j ≤ 4). However, this also changes states of all handles in row i and all handles in column j.

    The task is to determine the minimum number of handle switching necessary to open the refrigerator.

    Input

    The input contains four lines. Each of the four lines contains four characters describing the initial state of appropriate handles. A symbol “+” means that the handle is in closed state, whereas the symbol “−” means “open”. At least one of the handles is initially closed.

    Output

    The first line of the input contains N – the minimum number of switching. The rest N lines describe switching sequence. Each of the lines contains a row number and a column number of the matrix separated by one or more spaces. If there are several solutions, you may give any one of them.

    Sample Input

    -+--
    ----
    ----
    -+--

    Sample Output

    6
    1 1
    1 3
    1 4
    4 1
    4 3
    4 4

    受到POJ1753的启发,用枚举过了,但是这个题有高效的数学方法。。。

    #include <stdio.h>
    #include <queue>
    #include <string.h>
    using namespace std;
    
    struct node
    {
        int data, step;
    }path[65537];
    queue<struct node>q;
    
    int swit(int s, int n)
    {
        int t = n / 4 * 4;
        for(int i = t; i <= t+3; i++)
            s ^= 1<<i;
        for(int i = n % 4; i < 16; i += 4)
            s ^= 1<<i;
        s ^= 1<<n;
        return s;
    }
    
    int vis[65537];
    void bfs()
    {
        while(!q.empty())
        {
            struct node u = q.front();
            q.pop();
            if(u.data == 0)
            {
                printf("%d
    ", u.step);
                int tmp = u.data;
                for(int i = 0; i < u.step; i++)
                {
                    printf("%d %d
    ", 4-path[tmp].step/4, 4-path[tmp].step%4);
                    tmp = path[tmp].data;
                }
                return;
            }
            for(int i = 0; i < 16; i++)
            {
                int y = swit(u.data, i);
                if(!vis[y])
                {
                    q.push((struct node){y, u.step+1});
                    path[y] = (struct node){u.data, i};
                    vis[y] = 1;
                }
            }
        }
    }
    
    int main()
    {
        char s[5];
        int x = 0;
        for(int i = 0; i < 4; i++)
        {
            scanf("%s", s);
            for(int j = 0; j < 4; j++)
            {
                if(s[j] == '+')
                    x = (x<<1) + 1;
                else x <<= 1;
            }
        }
        memset(vis, 0, sizeof(vis));
        vis[x] = 1;
        q.push((struct node){x, 0});
        bfs();
        return 0;
    }
  • 相关阅读:
    交通综合改造工程EPC总承包项目
    二三维一体化地理信息平台
    NetCore3.1升级到Net5.0序列化方法过时问题
    windows server2012部署.net core IIS,页面报503,应用程序池自动停止。。。
    NetCore使用NPOI导入Word中的图片信息
    NetCore 使用 iTextSharp 读取 PDF 中的文字信息
    NetCore 在 Docker中文件路径找不到的问题
    Vue中数组list直接push的是对象而不是追加数据的问题
    netcore3.1增加阿里云OSS云存储服务
    Centos中Docker容器中程序访问宿主机Redis和Mysql
  • 原文地址:https://www.cnblogs.com/wolfred7464/p/3149885.html
Copyright © 2011-2022 走看看