D. Gargari and Permutations

D. Gargari and Permutations

time limit per test

2seconds

memory limit per test

256 megabytes

Gargari got bored to play with the bishops and now, after solving the problem about them, he is trying to do math homework. In a math book he have found k permutations. Each of them consists of numbers 1, 2, ..., n in some order. Now he should find the length of the longest common subsequence of these permutations. Can you help Gargari?

You can read about longest common subsequence there: https://en.wikipedia.org/wiki/Longest_common_subsequence_problem

Input

The first line contains two integers n and k (1 ≤ n ≤ 1000; 2 ≤ k ≤ 5). Each of the next k lines contains integers 1, 2, ..., n in some order — description of the current permutation.

Output

Print the length of the longest common subsequence.

 

input

4 3
1 4 2 3
4 1 2 3
1 2 4 3

output

3

Note

The answer for the first test sample is subsequence [1, 2, 3].

Solution

题意:现在给你K(2<=K<=5)个串,每个串都是1˜N 这N个数的全排列之一, 现在要你求出这K个串的最长公共子序列的长度

题解:假设我们已经找到了这个最长的公共子序列,我们开始分析这个公共子序列的性质.

　　 若x,y都是这个子序列中的元素,且x的位置在y的位置之前,那么我们可以知道,x,y 在给出的K个串中的位置也一定会满足id(x) < id(y) . 

　　所以就可以大力DP了....... 

　　F[i] = max(F[j] + 1 , F[i]) (id(j) <= id(i)) ...

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1000;
int a[9][maxn + 99],id[9][maxn + 99],n,k,f[maxn + 99];
int vis[maxn + 99];
inline bool check(int x,int y){
    for (int i = 2 ; i <= k ; ++i) if (id[i][x] > id[i][y]) return false;
    return true;
}
int main(){
    scanf("%d%d",&n,&k);
    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) id[i][a[i][j]] = j;

    for (int i = 1 ; i <= n ; ++i) f[i] = 1;
    for (int i = 1 ; i <= n ; ++i){
       for (int j = i + 1 ; j <= n ; ++j)
       if (check(a[1][i],a[1][j])) f[j] = max(f[j],f[i] + 1);
    }
    int ans = 0;
    for (int i = 1 ; i <= n ; ++i) if (f[i] > ans) ans = f[i];
    cout<<ans<<endl;
    return 0;
}