【CODEFORCES】 D. Gargari and Permutations

D. Gargari and Permutations

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

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].

题解：这一题听说可以用DAG解，但是我不会.....

我的想法是既然每个数都只会出现一次，那我就用一个数组S[I][J]记录第I层的数J在I+1层的位置，再用D[I]记录以A[I]结尾的时候最长的长度。可以看到，如果把每个相同的数连起来，最长序列中是不存在相交的线段的，所以我们就可以得到递推方程：

D[I]=max(D[J]+1,D[I])  （条件：数I的连线和J的连线不相交）

条件可以用数组S来判断，这样这个问题就成了“最长不下降子序列”的形状。最后找出D[I]的最大值即可。

#include <iostream>
#include <cstdio>

using namespace std;

int s[10][1002],n,k,i,ans,j,a[7][1002],d[1002];
int _check(int i,int j)
{
    int de=1,x=i,y=j;
    while (s[de][a[de][x]]>s[de][a[de][y]] && de<k)
    {
        x=s[de][a[de][x]];
        y=s[de][a[de][y]];
        de++;
    }
    if (de==k) return 1;
    else return 0;
}

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]);
            s[i-1][a[i][j]]=j;
        }
    for (int i=1;i<=n;i++)  d[i]=1;
    for (int i=1;i<=n;i++)
        for (int j=1;j<i;j++)
        if (_check(i,j)) d[i]=max(d[j]+1,d[i]);
    for (int i=1;i<=n;i++) ans=max(ans,d[i]);
    cout <<ans;
    return 0;
}