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
题目类型：简单字符处理+二分搜索
算法分析：由于n(10000)比m(10)大得多，所以可以考虑二重枚举m，然后依次枚举每一行，用map记录当前行的两个字符串的行标号并查找当前行的两个字符串是否出现过，若是则直接break后输出结果，复杂度是O(m^2nlogn)。注意两字符串若出现则只会出现一次并且查找后没有break会TLE，由于每次查找的是给定两列的行标，所以每次枚举行时需要清除dict，否则会WA
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/************************************************** filename :a.cpp author :maksyuki created time :2018/1/13 10:07:09 last modified :2018/1/13 11:11:08 file location :C:\Users\abcd\Desktop\TheEternalPoet ***************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("in", "r", stdin) #define CFO freopen ("out", "w", stdout) #define CPPFF ifstream cin ("in") #define CPPFO ofstream cout ("out") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC) #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7F7F7F7F; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1e4 + 66; const int maxnn = 88; int n, m, alen = 0; char s[maxn][16][88]; map<PII, int> data; map<string, int> dict; int getID(const char *s) { string tmp = s; if(!dict.count(tmp)) return dict[tmp] = alen++; else return dict[tmp]; } int main() { #ifdef LOCAL CFF; CFO; #endif while(scanf("%d %d", &n, &m) != EOF) { dict.clear(), data.clear(); alen = 0; char tmp[maxnn]; fgets(tmp, maxnn, stdin); for(int i = 0; i < n; i++) { fgets(tmp, maxnn, stdin); int cur = 0, len = 0; for(int j = 0; tmp[j]; j++) { if(tmp[j] == ',') { s[i][cur][len] = 0; cur++; len = 0; } else if(tmp[j] == '\r' || tmp[j] == '\n') { s[i][cur][len] = 0; break; } else s[i][cur][len++] = tmp[j]; } } bool is_pnf = true; int aa = -1, ab = -1, ba = -1, bb = -1; for(int i = 0; i < m && is_pnf; i++) for(int j = i + 1; j < m && is_pnf; j++) { data.clear(); for(int k = 0; k < n; k++) { int va = getID(s[k][i]), vb = getID(s[k][j]); if(!data.count(PII(va, vb))) data[PII(va,vb)] = k; else { is_pnf = false; aa = data[PII(va,vb)], ab = k; ba = i, bb = j; break; } } } if(is_pnf) puts("YES"); else { puts("NO"); printf("%d %d\n%d %d\n", aa + 1, ab + 1, ba + 1, bb + 1); } } return 0; } |