UVA - 1592 Database(枚举+map)

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Database
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	[email protected]
How to win ACM ICPC	Michael	[email protected]
Notes from ACM ICPC champion	Michael	[email protected]

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.

.
How to compete in ACM ICPC	1
How to win ACM ICPC	2
Notes from ACM ICPC champion	2
1	Peter [email protected]
2	Michael [email protected]

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 ≤ n ≤ 10000, 1 ≤ m ≤ 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 ha

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 ≤ r1, r2 ≤ n, r1 ̸= r2), on
the third line write two integer column numbers c1 and c2 (1 ≤ c1, c2 ≤ m, c1 ̸= c2), so that values in
columns c1 and c2 are the same in rows r1 and r2.

Simple Input
3 3
How to compete in ACM ICPC,Peter,[email protected]
How to win ACM ICPC,Michael,[email protected]
Notes from ACM ICPC champion,Michael,[email protected]
2 3
1,Peter,[email protected]
2,Michael,[email protected]

Simple Output
NO
2 3
2 3
YES

submit

题目大意：输入一个n行m列的数据库，寻找两个不同行和两个不同列 (即(r1,c1)和(r2,c1)相同，(r1,c2)和(r2,c2)相同.
思路：首先给字符串一个编号（map）便于字符串比较。如果四重循环枚举肯定超时，所以可以只枚举c1，c2，然后k从上到下扫描，用一个二元组map存两列的内容，如果map键值已经存在在这个二元组，映射到的就是r1，当前行即r2.
WA了n次，还有UVA不可以用取消同步。

附上AC代码：

#include
#include
#include
#include

using namespace std;
const int maxn=10000+5;
string s[maxn][15];//存字符串
map<string ,int>str_id;//将字符串转为id
typedef pair<int,int>p;
int n,m;

int main()
{
//    ios::sync_with_stdio(false);

    while(cin>>n>>m)
    {
       str_id.clear();
        getchar();
        //read
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                s[i][j]="";
                while(1)
                {
                    char c=getchar();
                    if(c==','||c=='\n')break;
                     s[i][j]+=c;
                }
                str_id[s[i][j]]=j*100000+i;//id
            }
        }
        //solve
        int flag=1;
        for(int i=1;ifor(int j=i+1;j<=m;j++)
            {
                mapint>dx;
                for(int k=1;k<=n;k++)
                {
                    int x1=str_id[s[k][i]],x2=str_id[s[k][j]];
                    p cnt=make_pair(x1,x2);
                    if(dx.count(cnt))
                    {
                        cout<<"NO"<cout<" "<cout<" "<0;
                        goto END;
                    }
                    dx[cnt]=k;
                }
            }
        }
        END:
            if(flag)cout<<"YES"<return 0;
}