uvaoj-1592:database_peter studies the theory of relational databases. -优快云博客

这篇博客探讨了数据库理论中的Peter Normal Form（PNF），解释了PNF的概念，即避免数据库表格中的数据冗余。文章通过举例说明了一个不遵循PNF的表格，并提出通过引入唯一标识符和拆分表格来解决这个问题。博主还分享了如何判断一个表格是否处于PNF的输入和输出格式，并提供了一种使用Map优化比较策略的解题思路。

Peter studies the theory of relational databases. Table in the relational database consists of values that are arranged in rows and columns.

There are different normal forms that database may adhere to. Normal forms are designed to minimize the redundancy of data in the database. For example, a database table for a library might have a row for each book and columns for book name, book author, and author's email.

If the same author wrote several books, then this representation is clearly redundant. To formally define this kind of redundancy Peter has introduced his own normal form. A table is in Peter's Normal Form (PNF) if and only if there is no pair of rows and a pair of columns such that the values in the corresponding columns are the same for both rows.

How to compete in ACM ICPC	Peter	peter@neerc.ifmo.ru
How to win ACM ICPC	Michael	michael@neerc.ifmo.ru
Notes from ACM ICPC champion	Michael	michael@neerc.ifmo.ru

The above table is clearly not in PNF, since values for 2rd and 3rd columns repeat in 2nd and 3rd rows. However, if we introduce unique author identifier and split this table into two tables -- one containing book name and author id, and the other containing book id, author name, and author email, then both resulting tables will be in PNF.

$\textstyle \parbox{.5\textwidth}{ \begin{center} \begin{tabular}{\vert l\vert l... ...\hline Notes from ACM ICPC champion & 2 \\ \hline \end{tabular} \end{center}}$ $\textstyle \parbox{.49\textwidth}{ \begin{center} \begin{tabular}{\vert l\vert ... ...ine 2 & Michael & michael@neerc.ifmo.ru \\ \hline \end{tabular} \end{center}}$

Given a table your task is to figure out whether it is in PNF or not.

Input

Input contains several datasets. The first line of each dataset contains two integer numbers n and m ( 1 $\le$ n $\le$ 10000, 1 $\le$ m $\le$ 10), the number of rows and columns in the table. The following n lines contain table rows. Each row has m column values separated by commas. Column values consist of ASCII characters from space (ASCII code 32) to tilde (ASCII code 126) with the exception of comma (ASCII code 44). Values are not empty and have no leading and trailing spaces. Each row has at most 80 characters (including separating commas).

Output

For each dataset, if the table is in PNF write to the output file a single word ``YES" (without quotes). If the table is not in PNF, then write three lines. On the first line write a single word ``NO" (without quotes). On the second line write two integer row numbers r₁ and r₂ ( 1 $\le$ r₁, r₂ $\le$ n, r₁ $\ne$ r₂), on the third line write two integer column numbers c₁ and c₂ ( 1 $\le$ c₁, c₂ $\le$ m, c1 $\ne$ c2), so that values in columns c₁and c₂ are the same in rows r₁ and r₂.

Sample Input

3 3
How to compete in ACM ICPC,Peter,peter@neerc.ifmo.ru
How to win ACM ICPC,Michael,michael@neerc.ifmo.ru
Notes from ACM ICPC champion,Michael,michael@neerc.ifmo.ru
2 3
1,Peter,peter@neerc.ifmo.ru
2,Michael,michael@neerc.ifmo.ru

Sample Output

NO
2 3
2 3
YES

题解：刘汝佳白书128页例题5-9；很明了，模拟题；

思路就是遍历，这里面不能直接遍历，需要一些技巧；

根据刘汝佳的方法，首先考虑利用map来把字符串转换为数字，将数字进行比较，这样可以将效率提升很大；

另外要把着c1,c2进行扫描，每一个c1,c2数对都对行进行扫描，从上至下；

详细代码如下：

code：

/*********************
2 3
1,Peter,peter@neerc.ifmo.ru
2,Michael,michael@neerc.ifmo.ru
3 3
How to compete in ACM ICPC,Peter,peter@neerc.ifmo.ru
How to win ACM ICPC,Michael,michael@neerc.ifmo.ru
Notes from ACM ICPC champion,Michael,michael@neerc.ifmo.ru
*********************/
#include <string>
#include <cstring>
#include <cstdio>
#include <map>
#include <vector>
#include <iostream>
using namespace std;
map<pair<int,int>,int> idcache;//扫描时需要用到的map；
map<string,int> cache;//给字符串编号是需要的map；

string str[100100];
int a[10010][11];
int cnt,num;
int n,m;

void Clear()//初始化；
{
    memset(a,0,sizeof a);
    idcache.clear();
    cache.clear();
    cnt=0;
    return ;
}

void read()
{
    string s;
    s.clear();
    num=0;
    for(int i=0; i<n; i++)
    {
        getline(cin,s);
        int len=s.size();
        for(int j=0; j<len; j++)
        {
            if(s[j]==',')
            {
                cnt++;
            }
            else str[cnt]+=s[j];
        }//上边这个循环将以‘，’为分隔符的字符串筛选出来；
        cnt++;
        for(int j=0; j<m; j++)
        {
            //cout<<str[j+i*m]<<endl;
            if(cache.count(str[j+i*m]))//出现过的字符串编相同的编号，没出现过的安排一个新的编号；
            {
                a[i][j]=cache[str[j+i*m]];
            }
            else
            {
                cache[str[j+i*m]]=num++;
                a[i][j]=num-1;
            }
            str[j+i*m].clear();//及时清理避免下次使用出错；
            //cout<<i<<" "<<j<<" "<<a[i][j]<<endl;
        }
    }
    return;
}

void print()
{
    int x,y;
    for(int i=0; i<m-1; i++)
    {
        for(int j=i+1; j<m; j++)
        {
            for(int k=0; k<n; k++)//以行为本对列进行遍历；
            {
                x=a[k][i],y=a[k][j];
                //cout<<x<<"   "<<y<<"   "<<endl;
                if(idcache.count(make_pair(x,y)))
                {
                    printf("NO\n");
                    printf("%d %d\n%d %d\n",idcache[make_pair(x,y)],k+1,i+1,j+1);
                    return;
                }
                else
                {
                    idcache[make_pair(x,y)]=k+1;
                }
            }
            idcache.clear();//避免不同行的相同字符串进行干扰；
        }
    }

cout<<"YES"<<endl;
return;
}

int main()
{
    while(cin>>n>>m)//row,col;
    {
        cin.get();
        Clear();
        read();
        print();
    }
    return 0;
}

笔记：细心，细心，再细心；

这道题从早上八点半卡到十一点，实在是有点可惜啊；Orz//。