【CF10D】LCIS（LCIS）

题意：求两个序列的LCIS

n,m<=300，a[i]<=1e9

题意：O(n^2)

O(n^3)的话设dp[i,j]为A终点为a【1..i】且B终点为b[j]的最大长度，分a[i]==b[j]和a[i]!=b[j]转移，枚举前一个在b中取的位置k转移

发现转移的下标集合每次只扩大最后一个，用前缀max保存

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef long double ld;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 typedef pair<ll,ll>P;
 #define N  510
 #define M  1000000
 #define INF 1e9
 #define fi first
 #define se second
 #define MP make_pair
 #define pb push_back
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1
 #define fors(i) for(auto i:e[x]) if(i!=p)

 const int MOD=1e9+,inv2=(MOD+)/;
       double eps=1e-;
       int dx[]={-,,,};
       int dy[]={,,-,};

 int dp[N][N],pre[N][N],a[N],b[N],c[N];

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 int main()
 {
     int n=read();
     rep(i,,n) a[i]=read();
     int m=read();
     rep(i,,m) b[i]=read();
     rep(i,,n)
     {
         int mx=,y=;
         rep(j,,m)
         {
             dp[i][j]=pre[i][j]=;
             if(a[i]!=b[j])
             {
                 dp[i][j]=dp[i-][j];
                 pre[i][j]=j;
             }
             if(a[i]>b[j]&&mx<dp[i-][j])
             {
                 mx=dp[i-][j];
                 y=j;
             }
             if(a[i]==b[j])
             {
                 dp[i][j]=mx+;
                 pre[i][j]=y;
             }
         }
     }
     int x=n,y,ans=;
     rep(i,,m)
      if(dp[n][i]>ans)
      {
          ans=dp[n][i];
          y=i;
      }
     printf("%d\n",ans);
     if(ans==) return ;
     int t=;
     while(x)
     {
         if(a[x]==b[y]) c[++t]=b[y];
         y=pre[x][y];
         x--;
     }
     per(i,ans,) printf("%d ",c[i]);
     return ;
 }