背包问题（01背包、完全背包）死记应试

最新推荐文章于 2021-08-01 09:37:50 发布

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最新推荐文章于 2021-08-01 09:37:50 发布

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分类专栏：数据结构与算法分析文章标签： dp 动态规划

本文链接：https://blog.youkuaiyun.com/Bing_Lee/article/details/77876037

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本文详细介绍了两种经典的背包问题：01背包和完全背包。通过具体的代码示例和伪代码，阐述了如何利用动态规划解决这两种问题，并给出了求解过程中状态转移方程的具体实现。

背包问题（01背包、完全背包）通用公式

01背包问题
有n件物品每件物品的数量只有一件，求在背包容积为m的情况所能装的物品价值最大是多少

#include<iostream> 
#include<algorithm> 
using namespace std;  

int main()  
{  
    int n,v;
    cout<<"please input number of goods:"<<endl<<"n = ";
    cin>>n;
    cout<<"please input volume of backpack:"<<endl<<"v = ";
    cin>>v;
    vector<int> value(n+1), room(n+1);
    for (int i = 1; i <= n; i++)
    {
        cout<<"please input good "<<endl<<i<<"'s room = ";
        cin>>room[i];
        cout<<"please input good "<<endl<<i<<"'s value = ";
        cin>>value[i];
    }

    vector<vector<int> > dp(n+1, vector<int>(v+1, 0));

    //Method 1:
    for (int i = 1; i <= n; ++i)
    {
        for (int j = 1; j <= v; ++j)
        {
            if (j - room[i] >= 0)
            {
                dp[i][j] = max(dp[i-1][j-room[i]] + value[i], dp[i-1][j]);
            }
            else
            {
                dp[i][j] = dp[i-1][j];
            }
        }
    }
    cout<<dp[n][v];
}

#include<iostream> 
#include<algorithm> 
using namespace std;  

int main()  
{  
    int n,v;
    cout<<"please input number of goods:"<<endl<<"n = ";
    cin>>n;
    cout<<"please input volume of backpack:"<<endl<<"v = ";
    cin>>v;
    vector<int> value(n+1), room(n+1);
    for (int i = 1; i <= n; i++)
    {
        cout<<"please input good "<<endl<<i<<"'s room = ";
        cin>>room[i];
        cout<<"please input good "<<endl<<i<<"'s value = ";
        cin>>value[i];
    }

    vector<int> dp(v+1, 0);

    //Method 2:
    for (int i = 1; i <= n; ++i)
    {
        for (int j = v; j >= room[i]; --j)
        {
        	dp[j] = max(dp[j-room[i]] + value[i], dp[j]);
        }
    }
    cout<<dp[v];
}

伪代码

dp[0...V]<-0
for i<- 1 to n
	for v <- V to w[i]
		dp[v] <- max{dp[v]; dp[v − w[i]] + v[i]}
out <- dp[m]

完全背包问题
条件与01背包相同，不同的是每件物品的数量都有无数件，求所装物品最大价值是多少

#include<iostream> 
#include<algorithm> 
using namespace std;  

int main()  
{  
    int n,v;
    cout<<"please input number of goods:"<<endl<<"n = ";
    cin>>n;
    cout<<"please input volume of backpack:"<<endl<<"v = ";
    cin>>v;
    vector<int> value(n+1), room(n+1);
    for (int i = 1; i <= n; i++)
    {
        cout<<"please input good "<<endl<<i<<"'s room = ";
        cin>>room[i];
        cout<<"please input good "<<endl<<i<<"'s value = ";
        cin>>value[i];
    }

    //Method 1:
    vector<vector<int> > dp(n+1, vector<int>(v+1, 0));
    for (int i = 1; i <= n; ++i)
    {
        for (int j = 1; j <= v; ++j)
        {
            for (int k = 0; k <= j / room[i]; ++k)
            {
                    dp[i][j] = max(dp[i][j], dp[i-1][j - k*room[i]]+k*value[i]);
            }
        }
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= v; j++)
        {
            cout<<dp[i][j]<<" ";
        }
        cout<<endl;
    }
}

#include<iostream> 
#include<algorithm> 
using namespace std;  

int main()  
{  
    int n,v;
    cout<<"please input number of goods:"<<endl<<"n = ";
    cin>>n;
    cout<<"please input volume of backpack:"<<endl<<"v = ";
    cin>>v;
    vector<int> value(n+1), room(n+1);
    for (int i = 1; i <= n; i++)
    {
        cout<<"please input good "<<endl<<i<<"'s room = ";
        cin>>room[i];
        cout<<"please input good "<<endl<<i<<"'s value = ";
        cin>>value[i];
    }

    vector<int> dp(v+1, 0);

    //Method 2:
    for (int i = 1; i <= n; ++i)
    {
        for (int j = room[i]; j <= v; ++j)
        {
        	dp[j] = max(dp[j-room[i]] + value[i], dp[j]);
        	cout<<dp[j]<<" ";
        }
        cout<<endl;
    }
}

通过伪代码来表示

dp[0...V]<-0
for i<- 1 to n
	for v <- w[i] to V
		dp[v] <- max{dp[v]; dp[v − w[i]] + v[i]
out <- dp[m]