Thursday, July 28, 2016

hihocoder Numeric Keypad - [微软2016校园招聘在线笔试第二场]


http://hihocoder.com/discuss/question/2807


The numberic keypad on your mobile phone looks like below:
1 2 3
4 5 6
7 8 9
  0
Suppose you are holding your mobile phone with single hand. Your thumb points at digit 1. Each time you can 1) press the digit your thumb pointing at, 2) move your thumb right, 3) move your thumb down. Moving your thumb left or up is not allowed.
By using the numeric keypad under above constrains, you can produce some numbers like 177 or 480 while producing other numbers like 590 or 52 is impossible.
Given a number K, find out the maximum number less than or equal to K that can be produced.

输入

The first line contains an integer T, the number of testcases.
Each testcase occupies a single line with an integer K.
For 50% of the data, 1 <= K <= 999.
For 100% of the data, 1 <= K <= 10500, t <= 20.

输出

For each testcase output one line, the maximum number less than or equal to the corresponding K that can be produced.
样例输入
3
25
83
131
样例输出
25
80
129

已知手机键盘为
1 2 3
4 5 6
7 8 9
  0
我们采用一种特殊的方式在手机上输入数字,即当按下一个键之后,只能把手指向下或者向右移动。开始时手指默认放在1上。 比如按下2之后,就不能再回到1这个按键。 问对于给定的数字S,能够通过这种特殊方式输入的最大不超过S的数字是多少。

http://www.ghosta3.com/blog/?p=87
根据题意,可以知道的是,按下某个键后,这个键左方及上方的键不能再用了。例如按下了[5]:
mic0201
所以我们可以得到一张表,以表示按下某个键后还剩下的合法按键:
 mic0202
 同时,我们模拟一下按键的过程,例如给出的数是k = 99099。
 [9][9][0]            //[0]不合法,且没有小于[0]的按键,回溯到第一次按[9]时
 [8][9][9][9][9]   //已知[9]不合法,按下[8],后面无论怎么按都不会大于k,
                          //所以后面全部按按下[8]后的最大合法按键[9]
到此,我们可以总结出一套算法:
1、从高位开始(数字的高位,数组中的低位,按字符串储存的话);
2、从不大于当前数位的合法按键中从大到小枚举,
      若有合法按键存在且等于当前数位,跳到下一位,重复2;
      若没有,则回溯;
      若小于当前数位,那么后面的所以数位则填充当前最大的合法值,结束;
3、重复2直到低位结束;

解题思路我们考虑如何模拟输入数字的过程:


从高位到低位,一位一位输入数字。
在已知S的前提下,我们要使得打出的数字尽可能接近S。则最好的方法就是从高位到低位,每一位都输入和S相同的数字。
先不考虑按键限制的情况下,可以写出一个函数:
dfs(depth):
    If depth >= length Then
        // 找到一个合法解
        // 由于我们总是从高到低对数字进行枚举,因此第一个找到的解一定是最大解
        Output(result)
        Return True
    End If
    For i = 9 .. 0
        If (i <= S[depth]) Then
            result[depth] = i
            If (dfs(depth+1)) Then
                Return True
            End If
        End If
    End For
    Return False
接下来考虑如何按键限制条件。
当我们输入一个数字后,只能再输入键盘上在其右边或下边的数字。由此我们可以得到一个转移矩阵g[10][10]
g[i][j]表示按下数字i后能否移动到数字j。若能够g[i][j]=true,否则g[i][j]=false
比如:
g[1][2] = true,g[2][1] = false
当我们输入了前一位之后,后一位可以使用哪些数字也就确定了,所以我们再参数中增加最后一个输入的数字last
在枚举数字时,我们也需要加入条件判断是否满足要求。
因此将函数改造为:
dfs(depth, last):
    If depth >= length Then
        // 找到一个合法解
        // 由于我们总是从高到低对数字进行枚举,因此第一个找到的解一定是最大解
        Output(result)
        Return True
    End If
    For i = 9 .. 0
        If (i <= S[depth] and go[last][i]) Then
            result[depth] = i
            If (dfs(depth+1, i)) Then
                Return True
            End If
        End If
    End For
    Return False
但这个函数现在还是有问题的。
由于在该函数中,我们保证了每一位数字一定小于等于S的对应位。
比如当S=300时,用我们这个函数找出的最优解是200,而实际上可以打出299这个数字。
也就是说,当满足某一个条件后,最优解某些数位上的数字是可能大于S对应位的。
这个条件是:当答案的其中一位已经小于S的对应位时,这一位之后的数位可以选择任意一个数字。
为了保证答案尽可能大,因此在后面的数位上我们要选择能够移动到的最大数字。
再一次改进后的函数为:
dfs(depth, last, below):
    If depth >= length Then
        // 找到一个合法解
        // 由于我们总是从高到低对数字进行枚举,因此第一个找到的解一定是最大解
        Output(result)
        Return True
    End If
    If below == True Then
        // 已经有高位小于S的对应位置
        //将后面的数位填充为last能够移动到的最大数字
        result[depth..length-1] = max{i|go[last][i]==true}
        Output(result)
        Return True
    End If
    For i = 9 .. 0
        If (i <= S[depth] and go[last][i]) Then
            result[depth] = i
            If (dfs(depth+1, i, i < S[depth])) Then
                Return True
            End If
        End If
    End For
    Return False
到此,给出的dfs函数即可通过所有的测试点。
由于本题仍然是多组输入数据,所以一定要记得初始化。
http://www.voidcn.com/blog/chenzhenyu123456/article/p-5031317.html
思路:DFS 从高位扫一遍,每次选尽可能大的数去填。

typedef long long LL;
typedef double DD;
bool g[10][10] =
{
    1, 0, 0, 0, 0, 0, 0, 0, 0, 0,
    1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
    1, 0, 1, 1, 0, 1, 1, 0, 1, 1,
    0, 0, 0, 1, 0, 0, 1, 0, 0, 1,
    1, 0, 0, 0, 1, 1, 1, 1, 1, 1,
    1, 0, 0, 0, 0, 1, 1, 0, 1, 1,
    0, 0, 0, 0, 0, 0, 1, 0, 0, 1,
    1, 0, 0, 0, 0, 0, 0, 1, 1, 1,
    1, 0, 0, 0, 0, 0, 0, 0, 1, 1,
    0, 0, 0, 0, 0, 0, 0, 0, 0, 1,
};
string str;
int ans[502];
bool DFS(int len, int lastval, int bit, bool yes)
{
    if(bit == len) return true;
    if(yes)
    {
        for(int i = bit; i < len; i++)
            ans[i] = lastval ? 9 : 0;
        return true;
    }
    int v = str[bit] - '0';
    for(int i = 9; i >= 0; i--)
    {
        if(g[lastval][i] && i <= v)
        {
            ans[bit] = i;
            if(DFS(len, i, bit+1, yes|i < v)) return true;
        }
    }
    return false;
}
int main()
{
    int t; cin >> t;
    while(t--)
    {
        cin >> str;
        int len = str.size();
        DFS(len, 1, 0, false);
        for(int i = 0; i < len; i++) cout << ans[i];
        cout << endl;
    }
    return 0;
}

https://github.com/MetooQ/hihoCoder/blob/master/numericKeypad.java

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