Title:
Find the longest common ascending subsequence
answer:
The longest common ascending subsequence = the longest common subsequence (LCS) and the longest ascending subsequence (LIS)
LCS core code:
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
if(a[i]==b[j])dp[i][j]=max(dp[i][j],dp[i-1][j-1]+1);
else dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
}
}
LIS core code:
for(int i=1;i<=n;i++)
{
dp[i]=1;
for(int j=1;j<i;j++)
{
if(a[j]<a[i])dp[i]=max(dp[i],dp[j]+1);
}
mx=max(mx,dp[i]);
}
The code for the longest common ascending subsequence is to find the longest ascending subsequence on the longest common subsequence. That is, on the premise of judging a[i] = = b[j], find the ascending sequence
f[i][j] to represent all a[1 ~ i] and b[1 ~ j] in which b[j ] The set of public ascending subsequences at the end; the value of
f[i][j] is equal to the maximum length of the subsequences of the set; the
complexity O(n 3 )
code:
for (int i = 1; i <= n; i ++ )
{
for (int j = 1; j <= n; j ++ )
{
f[i][j] = f[i - 1][j];
if (a[i] == b[j])
{
int maxv = 1;
for (int k = 1; k < j; k ++ )
if (a[i] > b[k])
maxv = max(maxv, f[i - 1][k] + 1);
f[i][j] = max(f[i][j], maxv);
}
}
}
Let's take a closer look at the code. The third level of for loop is used to find the length of the longest common ascending subsequence +1 that is less than a[i]. In fact, all we need to know is the longest common ascending subsequence before Length, so that we can use a variable val to store the maximum value in F[ i-1 ][ k ], so that the third layer of loop can be omitted
for (int i = 1; i <= n; i ++ )
{
int maxv = 1;
for (int j = 1; j <= n; j ++ )
{
f[i][j] = f[i - 1][j];
if (a[i] == b[j]) f[i][j] = max(f[i][j], maxv);
if (a[i] > b[j]) maxv = max(maxv, f[i - 1][j] + 1);
}
}
We can also reduce the dimension to one dimension.
We know that f[i][j] is derived from f[i-1][j].
We use f[i] to represent the first i elements of a sequence and b sequence ( LCIS) length, t is the position of the end element of the longest LCIS, the new transfer equation:
f[i] = f[t] +1 (a[i] = b[j])
This reduces to one dimension
for(int i=1;i<=n;i++){
maxx=0;
for(int j=1;j<=n;j++){
if(b[j]<a[i]&&maxx<f[j]) maxx=f[j];
if(b[j]==a[i]) f[j]=maxx+1;
}
}
Code:
Code for two-dimensional space:
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 3010;
int n;
int a[N], b[N];
int f[N][N];
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i ++ ) scanf("%d", &a[i]);
for (int i = 1; i <= n; i ++ ) scanf("%d", &b[i]);
for (int i = 1; i <= n; i ++ )
{
int maxv = 1;
for (int j = 1; j <= n; j ++ )
{
f[i][j] = f[i - 1][j];
if (a[i] == b[j]) f[i][j] = max(f[i][j], maxv);
if (a[i] > b[j]) maxv = max(maxv, f[i - 1][j] + 1);
}
}
int res = 0;
for (int i = 1; i <= n; i ++ ) res = max(res, f[n][i]);
printf("%d\n", res);
return 0;
}
Code for one-dimensional array:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<string>
#define ll long long
#define maxn 3050
#define inf 2147483647
#define mod 10003
#define eps 1e-6
#define pi acos(-1.0)
#define de(x) ((x)*(x))
using namespace std;
inline int read(){
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {
if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch)) {
x=x*10+ch-48;ch=getchar();}
return x*f;
}
int n,a[maxn],b[maxn],f[maxn],maxx;
signed main(){
n=read();
for(int i=1;i<=n;i++) a[i]=read();
for(int i=1;i<=n;i++) b[i]=read();
for(int i=1;i<=n;i++){
maxx=0;
for(int j=1;j<=n;j++){
if(b[j]<a[i]&&maxx<f[j]) maxx=f[j];
if(b[j]==a[i]) f[j]=maxx+1;
}
}
maxx=0;
for(int i=1;i<=n;i++) if(maxx<f[i]) maxx=f[i];
printf("%d",maxx);
return 0;
}