zoukankan      html  css  js  c++  java
  • 寒假集训——搜索 D

    #include <stdio.h>
    #include <stdlib.h>
    #include <iostream>
    #include <string.h>
    using namespace std;
    int num[4][10],n;
    int vis[1100],exa[10];
    
    void dfs(int ceng,int s)
    {
        vis[s]=1;
        if(ceng>=n) return;
        for(int i=1;i<=n;i++)
        {
            if(exa[i]) continue;
            for(int j=1;j<=6;j++)
            {
                exa[i]=1;
                dfs(ceng+1,s*10+num[i][j]);
                exa[i]=0;
            }
        }
    }
    
    int main()
    {
        while(scanf("%d",&n)!=EOF)
        {
            memset(vis,0,sizeof(vis));
            memset(exa,0,sizeof(exa));
            for(int i=1;i<=n;i++)
            {
               for(int j=1;j<=6;j++)
               {
                  scanf("%d",&num[i][j]);
               }
            }
            dfs(0,0);
            int mark;
            for(int i=1;i<1000;i++)
            {
                if(!vis[i])
                {
                    mark=i-1;
                    break;
                }
            }
            printf("%d
    ",mark);
        }
        return 0;
    }
    

      

    Absent-minded Masha got set of n cubes for her birthday.

    At each of 6 faces of each cube, there is exactly one digit from 0 to 9. Masha became interested what is the largest natural x such she can make using her new cubes all integers from 1 to x.

    To make a number Masha can rotate her cubes and put them in a row. After that, she looks at upper faces of cubes from left to right and reads the number.

    The number can't contain leading zeros. It's not required to use all cubes to build a number.

    Pay attention: Masha can't make digit 6 from digit 9 and vice-versa using cube rotations.

    Input

    In first line integer n is given (1 ≤ n ≤ 3) — the number of cubes, Masha got for her birthday.

    Each of next n lines contains 6 integers aij (0 ≤ aij ≤ 9) — number on j-th face of i-th cube.

    Output

    Print single integer — maximum number x such Masha can make any integers from 1 to x using her cubes or 0 if Masha can't make even 1.

    Examples

    Input
    3
    0 1 2 3 4 5
    6 7 8 9 0 1
    2 3 4 5 6 7
    Output
    87
    Input
    3
    0 1 3 5 6 8
    1 2 4 5 7 8
    2 3 4 6 7 9
    Output
    98

    Note

    In the first test case, Masha can build all numbers from 1 to 87, but she can't make 88 because there are no two cubes with digit 8.

    题目:D - Cubes for Masha
    思路:
    因为最多为三个魔方,意思是数字最大为999
    所以可以采用枚举遍历的方法。
    简单来说,就是用搜索把三个魔法可以组成的所有的数全部搜出来,然后找到最近的断点。
    具体:
    用一个二维数组,或者三个一维数组储存魔方上的六个数。
    用vis数组来表示这个数是否存在,1存在,0不存在。
    搜索里面,注意一个魔方不能同时用上面的两个数,所以要用东西标记这个魔方是否用过了

  • 相关阅读:
    java 包
    数据库查询操作练习
    solr全文检索实现原理
    前端页面设计问题小计
    送给自己的九封信
    bootstrap-table初使用
    bootstrap-treeview初使用
    windows:plsql配置oracle连接
    maven的安装和配置
    cxf+spring+restful简单接口搭建
  • 原文地址:https://www.cnblogs.com/EchoZQN/p/10337982.html
Copyright © 2011-2022 走看看