2020 GDUT Rating Contest I (Div. 2) A.Cow Gymnastics

来源 codeforces 2020 GDUT Rating Contest I (Div. 2)
题目：
A. Cow Gymnastics
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≤K≤10), Bessie ranks the N cows according to their performance (1≤N≤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
The first line of the input file contains two positive integers K and N. The next K lines will each contain the integers 1…N in some order, indicating the rankings of the cows (cows are identified by the numbers 1…N). If A appears before B in one of these lines, that means cow A did better than cow B.

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

Example
inputCopy
3 4
4 1 2 3
4 1 3 2
4 2 1 3
outputCopy
4
Note
The consistent pairs of cows are (1,4), (2,4), (3,4), and (1,3).

题意：计算有几对奶牛满足其中一只永远比另一只好。
思路：可以理解为图，每次practice中如果a牛>b牛，就理解为多一条a到b的单向路径，另开一二维数组b，b[i][j]表示i到j几条路径，如果刚好等于k，说明始终i牛大于j牛，n在20以内，直接用Floyd解。

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#define INF 0x3f3f3f3f
#define mod 1000000007
using namespace std;
int n,K,a[30][30],b[30][30],ans;int main()
{cin>>K>>n;for (int i=1;i<=K;i++)for (int j=1;j<=n;j++)scanf("%d",&a[i][j]);for (int i=1;i<=K;i++){for (int j=1;j<=n;j++){for (int k=j+1;k<=n;k++)b[a[i][j]][a[i][k]]++;}}for (int i=1;i<=n;i++)for (int j=1;j<=n;j++)if (b[i][j]>=K) ans++;cout<<ans;return 0;
}

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