QBXT DP & Graph Theory Class 2018.5.2 0/1 Backpack Special

PS: The title of this article comes from the courseware of Zhang Haowei (Zhang Guoyi) dalao of Peking University.

I.0/1 Backpack

There are n items, a knapsack of volume m, each item has a value vi, and a volume ti,
select a number of items such that the sum of the values is maximum when the sum of the volumes does not exceed m.
n<=1000, m<=10000.

Normal 0/1 backpack.

II.0/1 Backpack+

n <= 1000, vi <= 10。

According to the idea of memory search:

Let x=current item number, y=sum of current volume, z=sum of current value

When x and z are fixed, the smaller the y, the better. So take x and z as the state and y as the value of dp.

O(nvi)

III.0/1 Backpack++

n<=40

This question has nothing to do with DP. The correct solution is a halved search.

O(2^(n/2)*n) Search in half All items are divided into two halves, and the two halves are searched separately, a scheme <==> Take one on each side and put it together

IV.0/1 Backpack+++

sum of values modulo p max

n<=40 p<=10000

because (a+b+c)%p <==> ((a+b)%p+c)%p

So the value vi<=10000 at the same time

O (eg)

V.0/1 Backpack++++

For any ith item, find the maximum value when the ith item must not be in the backpack.
n<=1000, m<=10000.

The maximum value of an item k when it is not in the backpack is equivalent to dividing the maximum value into the maximum value of 1~k-1 and the maximum value of k+1~n.

We can calculate the maximum value of a prefix and the maximum value of a suffix in O(nm).

Then enumerate the volume of the following divisions and calculate ans.

For the first i items, the sum of volumes is j, and what is the maximum sum of values
A prefixed collection of items
suffix collection of items

If the xth item is gone, 1 ~x- 1   + x+ 1 ~ n and a suffix.
How much volume are allocated to each of them.
enumerate

Enumerate x, and then enumerate how much volume is assigned to the prefix, find max
f[i][j]   1 ~ i volume is the maximum value of j
g[i][j] i ~ n volume is the maximum value of j

for (x=1; x<=n; x++)
{
  ans=0;
  for (j=0; j<=m; j++)
    years =max(years,f[ x- 1 ][j]+g[x+ 1 ][m- j]);
  cout<<ans<<endl;
}

VI.0/1 Backpack +++++

Output Scheme
• n<=10000, m<=10000.
• TL=2s
• ML=1M

General 0/1 backpack output scheme

for (i=1; i<=n; i++)
{
  for (j=0; j<=m; j++)
  {
    dp[i][j]=dp[i-1][j];
    g[i][j]=j;
    if (j>=w[i])
      if (dp[i-1][j-w[i]]+v[i]>dp[i][j])
      {
        dp[i][j]=dp[i-1][j-w[i]]+v[i];
        g[i][j]=j-w[i];
      }
  }
}  // g[i][j] : dp[i][j] 由 dp[i-1][g[i][j]]转移得到
j=m;
for (i=n; i>=1; i--)
{
  if (j!= g[i][j]) the i-th item is selected
  j=g[i][j];
}

However, this question is very conscientious to set the memory to 1M, which means that we cannot open a two-dimensional array, that is, we can only use a one-dimensional array to store the solution.

Here zhw introduces a subtle approach:

g[i] means that dp[i][current volume v] is transferred from dp[i/2][g[i]], and then divides and conquers continuously.

The final segmented rectangle gets smaller and smaller, eventually forming a path from 1 to n.

In this way, the optimal solution is found within the space complexity of O(M).

QBXT DP & Graph Theory Class 2018.5.2 0/1 Backpack Special

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