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
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef unsigned int uint; 5 typedef unsigned long long ull; 6 typedef long double ld; 7 typedef pair<int,int> PII; 8 typedef pair<ll,ll> Pll; 9 typedef vector<int> VI; 10 typedef vector<PII> VII; 11 typedef pair<ll,ll>P; 12 #define N 510 13 #define M 1000000 14 #define INF 1e9 15 #define fi first 16 #define se second 17 #define MP make_pair 18 #define pb push_back 19 #define pi acos(-1) 20 #define mem(a,b) memset(a,b,sizeof(a)) 21 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++) 22 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--) 23 #define lowbit(x) x&(-x) 24 #define Rand (rand()*(1<<16)+rand()) 25 #define id(x) ((x)<=B?(x):m-n/(x)+1) 26 #define ls p<<1 27 #define rs p<<1|1 28 #define fors(i) for(auto i:e[x]) if(i!=p) 29 30 const int MOD=1e9+7,inv2=(MOD+1)/2; 31 double eps=1e-6; 32 int dx[4]={-1,1,0,0}; 33 int dy[4]={0,0,-1,1}; 34 35 int dp[N][N],pre[N][N],a[N],b[N],c[N]; 36 37 int read() 38 { 39 int v=0,f=1; 40 char c=getchar(); 41 while(c<48||57<c) {if(c=='-') f=-1; c=getchar();} 42 while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar(); 43 return v*f; 44 } 45 46 int main() 47 { 48 int n=read(); 49 rep(i,1,n) a[i]=read(); 50 int m=read(); 51 rep(i,1,m) b[i]=read(); 52 rep(i,1,n) 53 { 54 int mx=0,y=0; 55 rep(j,1,m) 56 { 57 dp[i][j]=pre[i][j]=0; 58 if(a[i]!=b[j]) 59 { 60 dp[i][j]=dp[i-1][j]; 61 pre[i][j]=j; 62 } 63 if(a[i]>b[j]&&mx<dp[i-1][j]) 64 { 65 mx=dp[i-1][j]; 66 y=j; 67 } 68 if(a[i]==b[j]) 69 { 70 dp[i][j]=mx+1; 71 pre[i][j]=y; 72 } 73 } 74 } 75 int x=n,y,ans=0; 76 rep(i,1,m) 77 if(dp[n][i]>ans) 78 { 79 ans=dp[n][i]; 80 y=i; 81 } 82 printf("%d ",ans); 83 if(ans==0) return 0; 84 int t=0; 85 while(x) 86 { 87 if(a[x]==b[y]) c[++t]=b[y]; 88 y=pre[x][y]; 89 x--; 90 } 91 per(i,ans,1) printf("%d ",c[i]); 92 return 0; 93 }