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 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].
思路：找出k個序列中都存在的最長的序列。那么這個串中的元素在這k個序列的相對位置都是相同的，那么我們找出在所有的數(shù)對(x,y)，這種數(shù)對的特性是在k個序列中x都在y的前面，這樣的序列，我們給他們建一條邊。然后記憶化搜索跑圖去得到一個最大值就可以了。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=1e6+100; const int maxm=1e3+100; struct edge{int to,next; }e[maxx<<1]; int a[maxx],head[maxx<<1],dp[maxm]; int n,m,tot;inline void init() {tot=0;memset(head,-1,sizeof(head));memset(dp,-1,sizeof(dp)); } inline void add(int u,int v) {e[tot].to=v,e[tot].next=head[u],head[u]=tot++; } inline int dfs(int u) {if(dp[u]!=-1) return dp[u];dp[u]=1;int sum=0;for(int i=head[u];i!=-1;i=e[i].next){int to=e[i].to;sum=max(sum,dfs(to));}dp[u]+=sum;return dp[u]; } int main() {scanf("%d%d",&n,&m);map<pair<int,int>,int> mp;init();for(int i=1;i<=m;i++){for(int j=1;j<=n;j++) scanf("%d",&a[j]);for(int x=1;x<=n;x++)for(int y=x+1;y<=n;y++) mp[make_pair(a[x],a[y])]++;}for(int i=1;i<=n;i++)for(int j=1;j<=n;j++) if(mp[make_pair(i,j)]==m) add(i,j);int _max=0;for(int i=1;i<=n;i++) _max=max(_max,dfs(i));cout<<_max<<endl;return 0; }

努力加油a啊，(^o)/~

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