01背包

最新推荐文章于 2018-06-11 08:54:00 发布

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#poj #01背包

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本文介绍了一个典型的背包问题案例，通过优化算法实现空间复杂度从O(nm)降至O(m)。文章提供了完整的代码示例，展示了如何在限定重量内选择最高价值的物品组合。

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示例题目：POJ3624
参考资料：背包九讲

Charm Bracelet
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 38631 Accepted: 16754
Description

Bessie has gone to the mall’s jewelry store and spies a charm bracelet. Of course, she’d like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi ≤ 400), a ‘desirability’ factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).

Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

Input

Line 1: Two space-separated integers: N and M
Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di

Output

Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

Sample Input

4 6
1 4
2 6
3 12
2 7
Sample Output

没有要求必须装满背包

MLE代码：
时空复杂度：O(nm)

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int c[4000],w[4000];
int dp[4000][13000];


int main()
{
    ios::sync_with_stdio(false);
    int n,m;
    while(cin>>n>>m)
    {
        for(int i=1;i<=n;i++)
        {
            cin>>c[i]>>w[i];
        }
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(j-c[i]>=0)
                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-c[i]]+w[i]);
                else dp[i][j]=dp[i-1][j];
            }
        }
        /*cout<<endl;
        for(int i=0;i<=n;i++)
        {
            for(int j=0;j<=m;j++)
            {
                cout<<dp[i][j]<<" ";
            }
            cout<<endl;
        }*/
        cout<<dp[n][m]<<endl;
    }
    return 0;
}

AC代码：
时间复杂度：O(nm)
空间复杂度：O(m)

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int c[4000],w[4000];
int dp[13000];

int main()
{
    ios::sync_with_stdio(false);
    int n,m;
    while(cin>>n>>m)
    {
        for(int i=1;i<=n;i++)
        {
            cin>>c[i]>>w[i];
        }
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            for(int j=m;j>=c[i];j--)
            {
                dp[j]=max(dp[j],dp[j-c[i]]+w[i]);
            }
        }
        cout<<dp[m]<<endl;
    }
    return 0;
}