POJ 2127 Greatest Common Increasing Subsequence

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 .

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【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

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【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--;    }}