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. How to compete in ACM ICPC 1 How to win ACM ICPC 2 Notes from ACM ICPC champion 2 1 Peter peter@neerc.ifmo.ru 2 Michael michael@neerc.ifmo.ru 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 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 ≤ 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. 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
#include<iostream> #include<cstdio> #include<vector> #include<string> #include<map> #include<sstream> using namespace std; typedef pair<int, int> PII; //pair是一个集合 const int maxr = 10000 + 5; const int maxc = 10 + 5; int m, n, db[maxr][maxc], cnt; map<string, int> id; int ID(const string& s) { //调用字符串 if (!id.count(s)) { id[s] = ++cnt; } return id[s]; } void find() { for (int c1 = 0; c1 < m; c1++) for (int c2 = c1 + 1; c2 < m; c2++) { //循环c1、c2列 map<PII, int> d; for (int i = 0; i < n; i++) { PII p = make_pair(db[i][c1], db[i][c2]); if (d.count(p)) {//检测map中的key值 printf("NO\n"); printf("%d %d\n", d[p] + 1, i + 1); printf("%d %d\n", c1 + 1, c2 + 1); return; } d[p] = i;//value值只要不是0就ok } } printf("YES\n"); } /* 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 */ int main() { string s; while (getline(cin, s)) { stringstream ss(s); if (!(ss >> n >> m)) break; //注意，这里是极好的一种判断输入有没有结束的方法，因为cin没有返回值（因为没有学过oop我对重载理解不深） cnt = 0; id.clear(); for (int i = 0; i < n; i++) { getline(cin, s); int lastpos = -1; for (int j = 0; j < m; j++) { int p = s.find(',', lastpos + 1);//第二个参数是开始寻找的位置 if (p == string::npos) p = s.length();//string:npos是个特殊值，说明查找没有匹配 //(lastpos + 1, p - lastpos - 1)这就是提取出来的一个字符串 db[i][j] = ID(s.substr(lastpos + 1, p - lastpos - 1));//引用这个单独的字符串 lastpos = p;//更新lastpos } } find(); } return 0; }代码还是挺不好懂的。。。加油吧
