【】 CF10D LCIS

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Longest common subsequence rise

\ (f [i] [j ] \) represents \ (A \) before \ (I \) the number of matching \ (B \) of \ (J \) number, and \ (B [j] \) longest common sub rises when sequence length Required

Transfer:

if(A[i]==B[j]) dp[i][j]=max(dp[i-1][k])+1; k=[1,2,...,j-1]，B[k]<B[j]=A[i]
else dp[i][j]=dp[i-1][j];

Recording at \ (dp [i-1] [j] \) maximum \ (B [K] <B [J] \) , the optimization \ (O (n ^ 2) \)

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

const int MAXN=510;

inline int read(){
    int x=0; char c=getchar();
    while(c<'0') c=getchar();
    while(c>='0') x=x*10+c-'0',c=getchar();
    return x;
}

int n,m,A[MAXN],B[MAXN];
int dp[MAXN][MAXN],g[MAXN][MAXN],pre[MAXN][MAXN];

inline void dfs(int i,int x){
    if(!x) return;
    dfs(i-1,pre[i][x]);
    if(pre[i][x]!=x)
        printf("%d ",B[x]);
}

int main()
{
    n=read();
    for(int i=1;i<=n;++i)
        A[i]=read();
    m=read();
    for(int i=1;i<=m;++i)
        B[i]=read();
    for(int i=1;i<=n;++i){
        int k=0;
        for(int j=1;j<=m;++j){
            dp[i][j]=dp[i-1][j];
            pre[i][j]=j;
            if(A[i]==B[j])
                dp[i][j]=dp[i-1][k]+1,pre[i][j]=k;
            if(B[j]<A[i]&&dp[i-1][j]>dp[i-1][k]) k=j;
        }
    }
    int Ans=0,k=0;
    for(int i=1;i<=m;++i)
        if(dp[n][i]>Ans) Ans=dp[n][i],k=i;
    printf("%d\n",Ans);
    dfs(n,k);
    return 0;
}