USACO 2019 December Contest, Bronze Problem 1. Cow Gymnastics

Category: 雷竞技ray官网下载 Date: 2022年7月1日上午11:46

In order to improve their physical fitness, the cows have taken up gymnastics! Farmer John designates his favorite cow Bessie to coach the $N$ other cows and to assess their progress as they learn various gymnastic skills.

In each of $K$ practice sessions ( $1 \leq K \leq 10$ ), Bessie ranks the $N$ cows according to their performance ( $1 \leq N \leq 20$ ). Afterward, she is curious about the consistency in these rankings. A pair of two distinct cows is consistent if one cow did better than the other one in every practice session.

Help Bessie compute the total number of consistent pairs.

INPUT FORMAT (file gymnastics.in):

The first line of the input file contains two positive integers $K$ and $N$ . The next $K$ lines will each contain the integers $1 \dots N$ in some order, indicating the rankings of the cows (cows are identified by the numbers $1 \dots N$ ). If $A$ appears before $B$ in one of these lines, that means cow $A$ did better than cow $B$ .

OUTPUT FORMAT (file gymnastics.out):

Output, on a single line, the number of consistent pairs.

SAMPLE INPUT:

SAMPLE OUTPUT:

The consistent pairs of cows are $(1, 4)$ , $(2, 4)$ , $(3, 4)$ , and $(1, 3)$ .

Problem credits: Nick Wu

USACO 2019 December Contest, Bronze Problem 1. Cow Gymnastics题解(雷竞技raybet安卓版下载提供，仅供参考)

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(Analysis by Nick Wu)

As the problem statement says, for each pair of cows $(A, B)$ , we will count how many practice sessions cow $A$ did better than cow $B$ in. If cow $A$ did better than cow $B$ in all $K$ practice sessions, we increment a counter, and we'll print out the value of the counter once we've looped over all pairs of cows.

In terms of implementation details, we can use a 2D array to store all of the rankings. Below is Brian Dean's code following this approach.

#include <iostream>
#include <fstream>
using namespace std;
 
int N, K;
int data[10][20];
 
bool better(int a, int b, int session)
{
  int apos, bpos;
  for (int i=0; i<N; i++) {
    if (data[session][i] == a) apos = i;
    if (data[session][i] == b) bpos = i;
  }
  return apos < bpos;
}
 
int Nbetter(int a, int b)
{
  int total = 0;
  for (int session=0; session<K; session++)
    if (better(a,b,session)) total++;
  return total;
}
 
int main(void)
{
  ifstream fin ("gymnastics.in");
  ofstream fout ("gymnastics.out");
  fin >> K >> N;
  for (int k=0; k<K; k++)
    for (int n=0; n<N; n++) 
      fin >> data[k][n];
  int answer = 0;
  for (int a=1; a<=N; a++)
    for (int b=1; b<=N; b++)
      if (Nbetter(a,b) == K) answer++;
  fout << answer << "\n";
  return 0;
}