POJ 2127 Greatest Common Increasing Subsequence_you are given two sequences of integer numbers. wr-优快云博客

本文链接：https://blog.youkuaiyun.com/Orion_Rigel/article/details/52279082

本文介绍了一种解决最长公共上升子序列问题的算法实现。通过动态规划方法，找到两个整数序列之间的最长公共且递增的子序列。文章提供了一个详细的C++实现案例。

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Description
You are given two sequences of integer numbers. Write a program to determine their common increasing subsequence of maximal possible length.
Sequence S1 , S2 , … , SN of length N is called an increasing subsequence of a sequence A1 , A2 , … , AM of length M if there exist 1 <= i1 < i2 < … < iN <= M such that Sj = Aij for all 1 <= j <= N , and Sj < Sj+1 for all 1 <= j < N .

【题目分析】
最长公共上升子序列问题，只需要巧妙地转换一下思路，就可最长公共子序列差不多了。三种情况的转移和处理都十分巧妙

【代码】

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
int n,m,a[5001],b[5001];
int dp[5001][5001],pre[5001][5001];
int out[5001];
int main()
{
//  freopen("sail.in","r",stdin);
//  freopen("sail.out","w",stdout);
    scanf("%d",&n);for (int i=1;i<=n;++i) scanf("%d",&a[i]);
    scanf("%d",&m);for (int i=1;i<=m;++i) scanf("%d",&b[i]);
    int ans=0,now,cnt=0;
    for (int i=1;i<=n;++i)
    {
        int k=0;
        for (int j=1;j<=m;++j)
        {
            if (a[i]!=b[j]) dp[i][j]=dp[i-1][j];
            if (a[i]>b[j]&&dp[i][j]>dp[i][k]) k=j;
            if (a[i]==b[j]) dp[i][j]=dp[i][k]+1,pre[i][j]=k;
        }
    }
//  for (int i=1;i<=n;++i)
//  {
//      for (int j=1;j<=m;++j)
//          cout<<dp[i][j]<<" ";
//      cout<<endl;
//  }
    int y=0,x=n;
    for (int i=1;i<=m;++i)
        if (ans<dp[n][i]) y=i,ans=dp[n][i];
    while (dp[x][y])
    {
        if (a[x]!=b[y]) x--;
        else out[++cnt]=a[x],y=pre[x][y];
    }
    cout<<cnt<<endl;
    while (cnt>0)
    {
        if (cnt==1) printf("%d",out[cnt]);
        else printf("%d ",out[cnt]);
        cnt--;
    }
}