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.
Examples
Input
4 31 4 2 34 1 2 31 2 4 3
Output
3
Note
The answer for the first test sample is subsequence [1, 2, 3].
OJ-ID：
CodeForce 113B
author：
Caution_X
date of submission：
2019-09-27
tags：
DP
description modelling：
求多个数列的LCS
major steps to solve it：
1.dp[i]:表示以数字i结尾得到的LCS，pos[i][j]表示数字j再第i个数列的位置，cnt[i]表示数字i出现了几次
2.从每个数列第一个数开始往后遍历，当cnt[i]=k时说明i可以作为LCS的一部分了
3.接下来需要讨论一下，在LCS中加入i对答案的影响
4.我们用vector<>存入所有可以作为LCS一部分的值，然后遍历vector中的数，判断二者的pos，来决定i应该插入在哪一个位置
5.遍历完成后vector<>加入i并且重新从2步骤开始
warnings：
AC code:
#include<bits/stdc++.h>
using namespace std;
int a[6][1010];
int dp[1010];//以i结尾的LCS
int cnt[1010],pos[6][1010];//pos[i][j]=：j在 i中出现的位置
vector<int> q;
int main()
{
//freopen("input.txt","r",stdin);
int n,k,ans=0;
scanf("%d%d",&n,&k);
for(int i=0;i<k;i++)
for(int j=0;j<n;j++)
scanf("%d",&a[i][j]);
memset(cnt,0,sizeof(cnt));
memset(dp,0,sizeof(dp));
for(int i=0;i<n;i++){
for(int j=0;j<k;j++){
int cur=a[j][i];
pos[j][cur]=i;
cnt[cur]++;
if(cnt[cur]==k){
if(q.empty()) dp[cur]=1;
else{
for(int kk=0;kk<q.size();kk++){
bool flag=false;
for(int l=0;l<k;l++){
if(pos[l][q[kk]]>pos[l][cur]){
flag=true;
break;
}
}
if(!flag) dp[cur]=max(dp[cur],dp[q[kk]]+1);
else dp[cur]=max(dp[cur],1);
}
}
ans=max(ans,dp[cur]);
q.push_back(cur);
}
}
}
printf("%d\n",ans);
return 0;
}
原文地址：https://www.cnblogs.com/cautx/p/11599511.html