Dynamic programming 01-LIS, LCS, digital triangle

LIS: the longest ascending subsequence problem

Boundary processing : f[0]=0;

Method 1 : Consider what states can be derived from f[i]

for(int i = 0; i < n; i++)  i:结尾下标
	for(int j = i + 1; j <= n; j++) 下一个数
		if(a[j] > a[i]) f[j]=max(f[j],f[i]+1);

Method 2 : Consider which states f[i] can be obtained from.
This method is more commonly used

for(int i = 1; i <= n; i++)
	for(int j = 0;j < i;j++) 前一个数 
		if(a[j] < a[i]) f[i]=max(f[i],f[j]+1); 

aims

 for(int i=1;i<=n;i++) ans=max(ans,f[i]);

What is the optimal solution?

	for(int i=1;i<=n;i++)
	{
    
    
		for(int j=0;j<i;j++)
		{
    
    
			if(a[j]>a[i])
				if(f[j]+1>f[i]){
    
    
					f[i]=f[j]+1;
					pos[i]=j;//记录由某点推导过来的该点的下标 
				}
		}
	}
	void print(int i)
	{
    
    
		if(i==0) return ;
		print(pos[i]);//递归处理 
		cout<<a[i]<<" ";
		 
	}
	
	int ans=0,tail;
	for(int i=1;i<=n;i++)
		if(f[i]>ans) ans=f[i],tail=i;
	cout<<ans<<endl;
	print(tail);

Dichotomous optimization

$q\left[i\right]$ represents the number at the end of the ascending subsequence of length i, the number in the enumerated sequence$a\left[i\right]$ , put it in$The$ largest in the$q$ array is less thana [i] a[i]After the number of$a$$[$$i$$]$ , it is realized by dichotomy. Time complexity$nlogn$ .

    for(int i=1;i<=n;i++) cin>>a[i];
    int len=0;
    for(int i=1;i<=n;i++)
    {
    
    
        int l=0,r=len;
        while(l<r)
        {
    
    
            int mid=(l+r+1)>>1;
            if(q[mid]<a[i]) l=mid;
            else r=mid-1;
        }
        len=max(len,r+1);
        q[r+1]=a[i];
    }
    cout<<len;

LCS: Longest Common Subsequence Problem

Border processing :

for(int i = 0; i <= n; i++) f[i][0]=0;
for(int j = 0; j <= m; j++) f[0][j]=0;

State transition

for(int i = 1;i <= n;i++)
	for(int j=1; j<= m ;j++)
		f[i][j]=max(f[i-1][j],f[i][j-1]);
		if(a[i] == a[j]) f[i][j]=max(f[i][j],f[i-1][j-1]+1);

aims

cout<<f[n][m]<<endl;

Digital triangle model

Idea The
current coordinates (i,j) can go to (i+1,j) (i+1,j+1)

Boundary processing

f[1][1]=a[1][1]

State transition

for(int i = 1;i<n;i++)
    for(int j=1;j<=i;j++){
    
    
        f[i+1][j]=max(f[i+1][j],f[i][j]+a[i+1][j]);
        f[i+1][j+1]=max(f[i+1][j+1],f[i][j]+a[i+1][j+1]);
    }

Target result

int ans=0;
for(int j=1;j<=n;j++)
    ans=max(ans,f[n][j]);