01背包,完全背包,多重背包模板及例题

本文深入解析了背包问题的各种类型,包括01背包、完全背包和多重背包的算法实现,提供了优化版本的代码示例,并附带具体例题解答,是理解和掌握背包问题算法的重要资源。

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    //每个物品的重量
    vector<int> weight;
    //每个物品的价值
    vector<int> value;
    //每个物品的数量
    vector<int> nums;
    //背包的总重量
    int all;
    //多少种物品
    int n;

01背包

一般版本

vector<vector<int>> dp(n + 1, vector<int>(all + 1, 0));
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= all; j++)
        {
            if(weight[i] <= j)
                dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);
            else
                dp[i][j] = dp[i - 1][j];
        }
    }

优化版本

vector<int> dp(all + 1, 0);
    for (int i = 1; i <= n; i++)
    {
        //写法一
        for (int j = all; j >= 1; j--)
        {
            if (weight[i] <= j)
                dp[j] = max(dp[j], dp[j - weight[i]] + value[i]);
        }
        //写法二
        for (int j = all; j >= weight[i]; j--)
        {
            dp[j] = max(dp[j], dp[j - weight[i]] + value[i]);
        }
    }

例题:https://vjudge.net/problem/HDU-1114

答案:

#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;


int main()
{
    int cases;
    cin >> cases;
    while(cases--)
    {
        //骨头n种,背包体积allv
        int n, allv;
        int tmp;
        cin >> n >> allv;
        vector<int> value(n + 1, 0);
        vector<int> volume(n + 1, 0);
        for (int i = 1; i <= n; i++)
        {
            cin >> tmp;
            value[i] = tmp;
        }
        for (int i = 1; i <= n; i++)
        {
            cin >> tmp;
            volume[i] = tmp;
        }
        vector<vector<int>> dp(n + 1, vector<int>(allv + 1, 0));
        for (int i = 1; i <= n; i++)
        {
            //注意从0开始
            for (int j = 0; j <= allv; j++)
            {
                if (volume[i] <= j)
                    dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - volume[i]] + value[i]);
                else
                    dp[i][j] = dp[i - 1][j];
            }
        }
        cout << dp[n][allv] << endl;
    }
    return 0;
}








int main()
{
    int cases;
    cin >> cases;
    while(cases--)
    {
        //骨头n种,背包体积allv
        int n, allv;
        int tmp;
        cin >> n >> allv;
        vector<int> value(n + 1, 0);
        vector<int> volume(n + 1, 0);
        for (int i = 1; i <= n; i++)
        {
            cin >> tmp;
            value[i] = tmp;
        }
        for (int i = 1; i <= n; i++)
        {
            cin >> tmp;
            volume[i] = tmp;
        }
        vector<int> dp(allv + 1, 0);
        for (int i = 1; i <= n; i++)
        {
            for (int j = allv; j >= volume[i]; j--)
            {
                dp[j] = max(dp[j], dp[j - volume[i]] + value[i]);
            }
        }
        cout << dp[allv] << endl;
    }
    return 0;
}

完全背包

一般版本

vector<vector<int>> dp(n + 1, vector<int>(all + 1, 0));
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= all; j++)
        {
            for (int k = 0; k * weight[i] <= j; k++)
            {
                dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - k * weight[i]] + k * value[i]);
            }
        }
    }

优化版本

vector<int> dp(n + 1, 0);
    for (int i = 1; i <= n; i++)
    {
        //写法一
        for (int j = 1; j <= all; j++)
        {
            if(weight[i] <= j)
                dp[j] = max(dp[j], dp[j - weight[i]] + value[i]);
        }
        //写法二
        for (int j = weight[i]; j <= all; j++)
        {
            dp[j] = max(dp[j], dp[j - weight[i]] + value[i]);
        }
    }

 

例题:https://vjudge.net/problem/HDU-2602

答案:

#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;

/*超时
int main()
{
    int cases;
    cin >> cases;
    while(cases--)
    {
        int all1, all2, all;
        cin >> all1 >> all2;
        all = all2 - all1;
        int INF = all2 * 1000;
        int n;
        cin >> n;
        vector<int> value(n + 1, 0);
        vector<int> weight(n + 1, 0);
        for (int i = 1; i <= n; i++)
        {
            cin >> value[i] >> weight[i];
        }
        vector<vector<int>> dp(n + 1, vector<int>(all + 1, INF));
        for (int i = 0; i <= n; i++)
        {
            dp[i][0] = 0;
        }
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= all; j++)
            {
                for (int k = 1; k * weight[i] <= j; k++)
                {
                    dp[i][j] = min(dp[i - 1][j], dp[i - 1][j - k * weight[i]] + k * value[i]);
                }
            }
        }
        int MIN = INF;
        for (int i = 1; i <= n; i++)
        {
            MIN = min(dp[i][all], MIN);
        }
        if(MIN >= INF)
            cout << "This is impossible." << endl;
        else
        {
            cout << "The minimum amount of money in the piggy-bank is " << MIN << "." << endl;
        }
        
    }
    return 0;
}
*/

int main()
{
    int cases;
    cin >> cases;
    while (cases--)
    {
        int all1, all2, all;
        cin >> all1 >> all2;
        all = all2 - all1;
        int INF = all2 * 1000;
        int n;
        cin >> n;
        vector<int> value(n + 1, 0);
        vector<int> weight(n + 1, 0);
        for (int i = 1; i <= n; i++)
        {
            cin >> value[i] >> weight[i];
        }
        vector<int> dp(all + 1, INF);
        dp[0] = 0;
        for (int i = 1; i <= n; i++)
        {
            for (int j = weight[i]; j <= all; j++)
            {
                dp[j] = min(dp[j], dp[j - weight[i]] + value[i]);
            }
        }
        int MIN = min(INF, dp[all]);
        if (MIN >= INF)
            cout << "This is impossible." << endl;
        else
        {
            cout << "The minimum amount of money in the piggy-bank is " << MIN << "." << endl;
        }
    }
    return 0;
}

多重背包

for (int i = 1; i <= n; i++)
    {
        while(nums[i] != 1)
        {
            weight.push_back(weight[i]);
            nums[i]--;
        }
    }
    vector<int> dp(weight.size() + 1, 0);
    for (int i = 1; i < weight.size(); i++)
    {
        for (int j = all; j >= weight[i]; j--)
        {
            dp[j] = max(dp[j], dp[j - weight[i]] + value[i]);
        }
    }

 

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