问题
This is a hard algorithms problem that :
Divide the list in 2 parts (sum) that their sum closest to (most) each other
list length is 1 <= n <= 100 and their(numbers) weights 1<=w<=250 given in the question.
For example : 23 65 134 32 95 123 34
1.sum = 256
2.sum = 250
1.list = 1 2 3 7
2.list = 4 5 6
I have an algorithm but it didn't work for all inputs.
- init. lists list1 = [], list2 = []
- Sort elements (given list) [23 32 34 65 95 123 134]
- pop last one (max one)
- insert to the list which differs less
Implementation : list1 = [], list2 = []
- select 134 insert list1. list1 = [134]
- select 123 insert list2. because if you insert to the list1 the difference getting bigger
3. select 95 and insert list2 . because sum(list2) + 95 - sum(list1) is less.
and so on...
回答1:
The problem is NPC, but there is a pseudo polynomial algorithm for it, this is a 2-Partition problem, you can follow the way of pseudo polynomial time algorithm for sub set sum problem to solve this. If the input size is related polynomially to input values, then this can be done in polynomial time.
In your case (weights < 250) it's polynomial (because weight <= 250 n => sums <= 250 n^2).
Let Sum = sum of weights, we have to create two dimensional array A, then construct A, Column by Column
A[i,j] = true if (j == weight[i] or j - weight[i] = weight[k] (k is in list)).
The creation of array with this algorithm takes O(n^2 * sum/2).
At last we should find most valuable column which has true value.
Here is an example:
items:{0,1,2,3} weights:{4,7,2,8} => sum = 21 sum/2 = 10
items/weights 0| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10
---------------------------------------------------------
|0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0
|1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0
|2 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1
|3 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1
So because a[10, 2] == true the partition is 10, 11
This is an algorithm I found here and edited a little bit to solve your problem:
bool partition( vector< int > C ) {
// compute the total sum
int n = C.size();
int N = 0;
for( int i = 0; i < n; i++ ) N += C[i];
// initialize the table
T[0] = true;
for( int i = 1; i <= N; i++ ) T[i] = false;
// process the numbers one by one
for( int i = 0; i < n; i++ )
for( int j = N - C[i]; j >= 0; j--)
if( T[j] ) T[j + C[i]] = true;
for(int i = N/2;i>=0;i--)
if (T[i])
return i;
return 0;
}
I just returned first T[i] which is true instead of returning T[N/2] (in max to min order).
Finding the path which gives this value is not hard.
回答2:
You can reformulate this as the knapsack problem.
You have a list of items with total weight M that should be fitted into a bin that can hold maximum weight M/2. The items packed in the bin should weigh as much as possible, but not more than the bin holds.
For the case where all weights are non-negative, this problem is only weakly NP-complete and has polynomial time solutions.
A description of dynamic programming solutions for this problem can be found on Wikipedia.
回答3:
This problem is at least as hard as the NP-complete problem subset sum. Your algorithm is a greedy algorithm. This type of algorithm is fast, and can generate an approximate solution quickly but it cannot find the exact solution to an NP-complete problem.
A brute force approach is probably the simplest way to solve your problem, although it is will be to slow if there are too many elements.
- Try every possible way of partitioning the elements into two sets and calculate the absolute difference in the sums.
- Choose the partition for which the absolute difference is minimal.
Generating all the partitions can be done by considering the binary representation of each integer from 0 to 2^n, where each binary digit determines whether the correspending element is in the left or right partition.
来源：https://stackoverflow.com/questions/4479479/divide-list-in-two-parts-that-their-sum-closest-to-each-other