zoukankan      html  css  js  c++  java
  • cf 463D

    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].

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<string>
    #include<cmath>
    #include<cstdlib>
    #include<algorithm>
    using namespace std;
    int n,k,cnt[1002][1002],a[1002],f[1002];
    int main()
    {
          int ans=0;
          scanf("%d%d",&n,&k);
          for(int i=0;i<k;i++)
          {
                for(int j=0;j<n;j++)
                {
                      scanf("%d",&a[j]);
                      for(int c=0;c<j;c++)
                            cnt[a[c]][a[j]]++;
                }
          }
          for(int i=0;i<n;i++)
          {
               f[a[i]]=1;
               for(int j=0;j<i;j++)
               {
                     if(cnt[a[j]][a[i]]==k)
                     {
                           f[a[i]]=max(f[a[i]],f[a[j]]+1);
                     }
               }
               ans=max(ans,f[a[i]]);
          }
          printf("%d
    ",ans);
          return 0;
    }
    

      

  • 相关阅读:
    MFC通过Http Post数据到Web端
    C++解析JSON格式数据
    APScheduler最基本的用法
    error connection reset by peer 104
    navicat远程连接mysql错误
    ubuntu18.04 校准时间
    ubuntu下python在pycharm环境下安装setuptools和pip,和distutils.core
    ubuntu下pycharm快捷方式创建
    django无法加载样式
    YAML快速入门
  • 原文地址:https://www.cnblogs.com/a972290869/p/4240708.html
Copyright © 2011-2022 走看看