标签：并且 应该 mes def ack turn can mem 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
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.
OutputPrint the length of the longest common subsequence.
Examples4 3
1 4 2 3
4 1 2 3
1 2 4 3
3
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; }
标签：并且 应该 mes def ack turn can mem output
原文地址：https://www.cnblogs.com/cautx/p/11599511.html
