matlab拔河比赛_拔河比赛

matlab拔河比赛

Description:

描述：

This is a standard interview problem to divide a set of number to two different set where both the subset contains same number of element and have a minimum difference of sum between them using backtracking.

这是一个标准的采访问题，它是将一组数字划分为两个不同的集合，其中两个子集都包含相同数量的元素，并且使用回溯，它们之间的总和差异最小。

Problem statement:

问题陈述：

There is a set contains N number of elements. We have to divide this set into two different sets where each of the subsets contains the same number of elements and has a minimum difference between them.

有一个包含N个元素的集合。我们必须将此集合分为两个不同的集合，其中每个子集包含相同数量的元素，并且它们之间的差异最小。

If N is even then each of the subsets will contain N/2 number of elements.
如果N为偶数，则每个子集将包含N / 2个元素。
If N is odd then one of the subsets will contain (N+1)/2 and others will contain (N-1)/2 number of elements.
如果N为奇数，则子集中的一个子集将包含(N + 1)/ 2个元素，而其他子集将包含(N-1)/ 2个元素。

    Input:
Test case T
//T no. of line with the value of N and corresponding values.
E.g.
3
3
1 2 3
4
1 2 3 4
10
1 4 8 6 -7 -10 87 54 16 100
1<=T<=100
1<=N<=100
Output:
Print the two subsets.

Example

例

    T=3
N=3
1 2 3
Output:  set1: {1, 2} set2: {3}
N=4
1 2 3 4
Output: set1: {1, 4} set2: {2, 3}
N=10
1 4 8 6 -7 -10 87 54 16 100
Output: set1: {{ 4, -7, -10, 87, 54 } set2: {1, 8, 6, 16, 100 }

Explanation with example

举例说明

Let, there is N no. of values, say e₁, e₂, e₃, ..., e_n and N is even.

让我们没有N。 e ₁ ，e ₂ ，e ₃ ，...，e _n和N的值是偶数。

    Subset1: { N/2  no. of values }
Subset2: { N/2 no. of values }

The process to insert the elements to the Subsets is a problem of combination and permutation. To get the result we use the backtracking process. Here we take the sum of all the elements

. Here we have to find out the elements that are in the subset no.1 and the elements that are not in subset1, must be in the subset no. 2. Every time we consider two things.

将元素插入子集的过程是组合和排列的问题。为了获得结果，我们使用了回溯过程。这里我们取所有元素的总和

The i^th element is a part of the subset1.
第i ^个元素是subset1的一部分。
The i^th element is not a part of the subset1.
第i ^个元素不是subset1的一部分。

After getting N/2 number of elements into the subset1. We will check,

在将N / 2数量的元素放入子集1之后。我们会检查

If the value is less than then Difference we update the value of the Difference and make a note of the elements that are in the subset1.

如果该值小于那么差，我们更新之差的值并记录是在SUBSET1元素。

After getting the elements for those the value of Difference is less we check the elements of the main set and find out the elements that are not in the subset1 and put them in subset2.

获得元素对于那些经过差值小于我们检查的主要集合中的元素，并找出不在SUBSET1的元素，并把它们放在SUBSET2。

    Set: {1, 2, 3}
Sum=6

1 is in the subset1 therefore Subset1: {1} and Difference = abs(3-1) = 2
1在子集1中，因此子集1：{1}，并且差= abs(3-1)= 2
2 is in the subset1 therefore Subset1: {1,2} and Difference = abs(3-3) = 0
2在子集1中，因此子集1：{1,2}，且差= abs(3-3)= 0

Therefore, Difference is minimum.

因此，差异最小。

    Subset1: {1,2}
Subset2: {3}

C++ implementation:

C ++实现：

#include <bits/stdc++.h>
using namespace std;
void traverse(int* arr, int i, int n, int no_of_selected, int sum, int& diff, bool* take, vector<int>& result, vector<int> current, int current_sum)
{
//if the current position is greater than or equals to n
if (i >= n)
return;
//if the value of difference is greater than
//Sum/2-current sum of the elements of Subset no.1
if ((diff > abs(sum / 2 - current_sum)) && (no_of_selected == (n + 1) / 2 || no_of_selected == (n - 1) / 2)) {
diff = abs(sum / 2 - current_sum);
//store the subset no. 1
result = current;
}
//taking the elemets after the current element one by one
for (int j = i; j < n; j++) {
take[j] = true;
current.push_back(arr[j]);
traverse(arr, j + 1, n, no_of_selected + 1, sum, diff, take, result, current, current_sum + arr[j]);
current.pop_back();
take[j] = false;
}
}
void find(int* arr, int n)
{
//array to distinguished the elements those are in subset no. 1
bool take[n];
int sum = 0;
for (int i = 0; i < n; i++) {
sum += arr[i];
take[i] = false;
}
int diff = INT_MAX;
int no_of_selected = 0;
vector<int> result;
vector<int> current;
int current_sum = 0;
int i = 0;
traverse(arr, i, n, no_of_selected, sum, diff, take, result, current, current_sum);
set<int> s;
//elements those are in subset no.1
cout << "Set1 : { ";
for (int j = 0; j < result.size(); j++) {
cout << result[j] << " ";
s.insert(result[j]);
}
cout << "}" << endl;
//elements those are in subset no.2
cout << "Set2 : { ";
for (int j = 0; j < n; j++) {
if (s.find(arr[j]) == s.end()) {
cout << arr[j] << " ";
}
}
cout << "}" << endl;
}
int main()
{
int t;
cout << "Test case : ";
cin >> t;
while (t--) {
int n;
cout << "Enter the value of n : ";
cin >> n;
int arr[n];
cout << "Enter the values: ";
//taking the set elements
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
find(arr, n);
}
return 0;
}

Output

输出量

Test cases : 3
Enter the value of n : 3
Enter the values: 1 2 3
Set1 : { 1 2 }
Set2 : { 3 }
Enter the value of n : 4
Enter the values: 1 2 3 4
Set1 : { 2 3 }
Set2 : { 1 4 }
Enter the value of n : 10
Enter the values: 1 4 8 6 -7 -10 87 54 16 100
Set1 : { 4 -7 -10 87 54 }
Set2 : { 1 8 6 16 100 }