A - Charm Bracelet（0 1背包）

 

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 W_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 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: W_i and D_i

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

每个只用一次，0，1背包模板题

#include<iostream>
#include<cstring>
using namespace std;
const int maxn=1e5+1000;
int n,m;
int w[maxn],d[maxn];
int dp[maxn];
int main()
{
    cin>>n>>m;
    memset(dp,0,sizeof(dp));
    for(int i=0;i<n;i++)
    {
        cin>>w[i]>>d[i];
    }
    for(int i=0;i<n;i++)
    {
        for(int j=m;j>=w[i];j-- )
        {
            dp[j]=max(dp[j],dp[j-w[i]]+d[i]);
        }
    }
    cout<<dp[m]<<endl;
    return 0;
} 

选出最大，一次一次更新，

dp[j]=max(dp[j],dp[j-w[i]]+d[i])

用二维数组来写开1e6会炸，

    for(int i=1;i<=n;i++)
      for(int j=0;j<=m;j++)
      {
          dp[i][j]=dp[i-1][j];//不包括第i个小于j最大的值为dp[i-1][j]

          if(j>=w[i])//在范围内
            dp[i][j]=max(dp[i][j],dp[i-1][j-w[i]]+d[i]);
      }
    cout<<dp[n][m]<<endl;

转换成一维数组，用滚动数组，就可以了

  for(int i=0;i<n;i++)
    {
        for(int j=m;j>=w[i];j-- )
        {
            dp[j]=max(dp[j],dp[j-w[i]]+d[i]);
        }
    }