01背包

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

23

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;}