Dynamic planning: code summary of knapsack problem template

explain ： according to acwing Algorithm improvement course and algorithm foundation course , The code is based on y In general, slightly modified .

1.0-1 knapsack problem

Problem model description ：

Yes  N  Items and a capacity are  V  The backpack . Each item can only be used once .

The first  i  The volume of the item is vi, The value is  wi. Find out which items to pack in your backpack , The total volume of these items can not exceed the capacity of the backpack , And the total value is the greatest .

The template questions ：2. 01 knapsack problem - AcWing Question bank
Code （ Space optimized version ）：

#include<bits/stdc++.h>
using namespace std;
const int N=1010;
int f[N];
int main()
{
	int n,m;
	cin>>n>>m; // Quantity and total volume  
	for(int i=0;i<n;++i){
		int v,w; //v: Volume ;w: value  
		cin>>v>>w;
		for(int j=m;j>=v;--j) f[j]=max(f[j],f[j-v]+w);
	}
	cout<<f[m];
	return 0;
}

 2. Complete knapsack problem

Problem model description ：0-1 Based on the knapsack problem , Every item can be selected indefinitely .

The template questions ：3. Complete knapsack problem - AcWing Question bank

Code （ Optimize to O(n^2), One dimensional space ）：

#include<bits/stdc++.h>
using namespace std;
const int N=1010;
int f[N];
int main()
{
	int n,m;
	cin>>n>>m; // Quantity and total volume  
	for(int i=0;i<n;++i){
		int v,w; //v: Volume ;w: value  
		cin>>v>>w;
		for(int j=v;j<=m;++j) f[j]=max(f[j],f[j-v]+w);
	}
	cout<<f[m];
	return 0;
}

3. Group knapsack problem  

Problem model description ：

Yes  N  Group items and a capacity are  V  The backpack . There are several items in each group , Only one item can be selected in the same group .

Find out which items to pack in your backpack , It can make the total volume of the goods not exceed the capacity of the backpack , And the total value is the greatest .

The template questions ：9. Group knapsack problem - AcWing Question bank

Code ：

#include<bits/stdc++.h>
using namespace std;
const int N=110;
int f[N],v[N][N],w[N][N],s[N];
int main()
{
	int n,m;
	cin>>n>>m; // Quantity and total volume  
	for(int i=1;i<=n;++i){ 
		cin>>s[i]; // The first i Number of classes  
		for(int j=1;j<=s[i];++j) cin>>v[i][j]>>w[i][j]; 
	}
	for(int i=1;i<=n;++i)
	for(int j=m;j>=0;--j)
	for(int k=1;k<=s[i];++k)
	if(v[i][k]<=j) f[j]=max(f[j],f[j-v[i][k]]+w[i][k]);
	cout<<f[m];
	return 0;
}

4. The longest common ascending subsequence problem

This is not a backpack model , It's an interval dp, The complexity of simple practice is O(n^3), Put it here O(n^2) Complexity code .

Problem model description ： Two sequences  A  and  B, If they all contain a number whose position is not necessarily continuous , And the value is strictly increasing , Then we call this segment a common ascending subsequence of two sequences , And the longest of all the common ascending subsequences is the longest common ascending subsequence .

The template questions ：272. The longest common ascending subsequence - AcWing Question bank

Code （O(n^2)）：

#include<bits/stdc++.h>
using namespace std;
const int N=3010;
int n,a[N],b[N],f[N][N];
int main()
{
    cin>>n;
    for(int i=1;i<=n;++i) cin>>a[i];
    for(int i=1;i<=n;++i) cin>>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(b[j]<a[i]) maxv=max(maxv,f[i][j]+1);
        }
    }
    int res=0;
    for(int i=1;i<=n;++i) res=max(res,f[n][i]);
    cout<<res;
    return 0;
}

5. The knapsack problem is the number of solutions

Problem model description ：0-1 Based on the knapsack problem , Number of schemes of the most optimal method .

The template questions ：11. The knapsack problem is the number of solutions - AcWing Question bank

Code ：

#include<bits/stdc++.h>
using namespace std;
const int N=1010,mod=1e9+7;
int f[N],g[N];
int main()
{
    int n,m;
    cin>>n>>m;
    g[0]=1;
    for(int i=0;i<n;++i){
        int v,w;
        cin>>v>>w;
        for(int j=m;j>=v;--j){
            int mav=max(f[j],f[j-v]+w),cnt=0;
            if(mav==f[j]) cnt+=g[j];
            if(mav==f[j-v]+w) cnt+=g[j-v];
            cnt%=mod;
            g[j]=cnt;
            f[j]=mav;
        }
    }
    int res=0,ans=0;
    for(int i=0;i<m;++i) res=max(res,f[i]);
    for(int i=0;i<m;++i) if(res==f[i]) ans=(ans+g[i])%mod;
    cout<<ans;
    return 0;
}

6. There is a knapsack problem of dependence .

Problem model description ： Yes  N  One item and one capacity are  V  The backpack . There is a dependency between objects , And dependencies make up the shape of a tree . If you choose an item , You must select its parent node .

The template questions ：10. There is a knapsack problem of dependence - AcWing Question bank

State means ：f[u][m]: With u The subtree of the root node , At a cost of no more than m The maximum value generated by .

#include<bits/stdc++.h>
using namespace std;
const int N=110;
int h[N],ne[N],e[N],idx,v[N],w[N],f[N][N];
int n,m;
void add(int a,int b)
{
	e[idx]=b,ne[idx]=h[a],h[a]=idx++;
}
void dfs(int u)
{
	for(int i=h[u];~i;i=ne[i]){
		int son=e[i];
		dfs(son);
		for(int j=m-v[u];j>=0;--j)
		for(int k=0;k<=j;++k)
		f[u][j]=max(f[u][j],f[u][j-k]+f[son][k]);
	}
	for(int i=m;i>=v[u];--i) f[u][i]=f[u][i-v[u]]+w[u];
	for(int i=0;i<v[u];++i) f[u][i]=0;
}
int main()
{
	cin>>n>>m;
	memset(h,-1,sizeof(h));
	int r;
	for(int i=1;i<=n;++i){
		int p;
		cin>>v[i]>>w[i]>>p;
		if(p==-1) r=i;
		else add(p,i);
	}
	dfs(r);
	cout<<f[r][m];
	return 0;
}

 7. Multiple knapsack problem

Problem model description ：0-1 Based on the knapsack problem , The first i Choose at most items si Pieces of .

The template questions ：5. Multiple knapsack problem II - AcWing Question bank

Code （ Binary optimized version ）：

#include<bits/stdc++.h>
using namespace std;
const int M=2010,N=25000;
int v[N],w[N],f[M];
int main()
{
    int n,m;
    cin>>n>>m;
    int cnt=0;
    for(int i=1;i<=n;++i){
    	int a,b,s;
    	cin>>a>>b>>s;
    	int k=1;
    	while(k<=s){
    		++cnt;
    		v[cnt]=a*k,w[cnt]=b*k;
    		s-=k;
    		k<<=1;
		}
		if(s>0){
			++cnt;
			v[cnt]=a*s,w[cnt]=b*s;
		}
	}
	n=cnt;
	for(int i=1;i<=n;++i)
	for(int j=m;j>=v[i];--j)
	f[j]=max(f[j],f[j-v[i]]+w[i]);
	cout<<f[m];
    return 0;
}

Sliding window optimization ：

#include<bits/stdc++.h>
using namespace std;
const int N=20010;
int f[N],g[N],q[N];
int main()
{
    int n,m;
    cin>>n>>m;
    for(int i=0;i<n;++i){
        int v,w,s;
        cin>>v>>w>>s;
        memcpy(g,f,sizeof(f)); // Save one level of status 
        for(int j=0;j<v;++j){
            int hh=0,tt=-1;
            for(int k=j;k<=m;k+=v){
                if(hh<=tt&&q[hh]<k-s*v) hh++;
                if(hh<=tt) f[k]=max(f[k],g[q[hh]]+(k-q[hh])/v*w);
                while(hh<=tt&&g[q[tt]]-(q[tt]-j)/v*w<=g[k]-(k-j)/v*w) tt--;
                q[++tt]=k;
            }
        }
    }
    cout<<f[m];
    return 0;
}

 

 8. Combinatorial knapsack problem

Problem model description ： Yes  N  Items and a capacity are  V  The backpack .

There are three kinds of goods ：

• The first kind of goods can only be used 1 Time （01 knapsack ）;
• The second kind of goods can be used indefinitely （ Completely backpack ）;
• The third kind of goods can only be used at most  si  Time （ Multiple backpack ）;

The template questions ：7. Mixed knapsack problem - AcWing Question bank

Code ：

#include<bits/stdc++.h>
using namespace std;
const int N=1010;
int n,m;
int f[N];
int main()
{
    cin>>n>>m;
    for(int i=0;i<n;++i){
        int v,w,s;
        cin>>v>>w>>s;
        if(s==0) // Completely backpack 
        for(int j=v;j<=m;++j) f[j]=max(f[j],f[j-v]+w);
        else{
            if(s==-1) s=1; //01 The number of backpacks is 1 Multiple backpacks 
            for(int k=1;k<=s;k*=2){
                for(int j=m;j>=k*v;--j) f[j]=max(f[j],f[j-k*v]+k*w);
                s-=k;
            }
            if(s) for(int j=m;j>=s*v;--j) f[j]=max(f[j],f[j-s*v]+s*w);
        }
    }
    cout<<f[m];
    return 0;
}