[YBT high-efficiency advanced] 1 basic algorithm/4 depth-first search/1 tug of war

[YBT high-efficiency advanced] 1 basic algorithm/4 depth-first search/1 tug of war

Memory limit: 256 MiB
Time limit: 1000 ms
standard input and output
Question type: Traditional
Evaluation method: Text comparison

Title description

In the ACM training team of Zhejiang Normal University, the team members usually have a heavy mental and labor force. In order to combine work and rest, our beloved Teacher Han prepared a tug-of-war competition to let the team members relax.

For the fairness of the tug-of-war competition, Mr. Han puts forward the following requirements:
1. The maximum number of people on both sides of the tug-of-war competition cannot differ by $1$ .
2. Every team member has a weight, and we want to minimize the weight and difference between the two players.

Now has $For\; N$ players, Mr. Han would like you to help distribute them, and output the minimum value of the difference between the weights of the two sides after the distribution.

Input format

First enter $T$ , means there isTTExample of group$T.$

Each example:

First enter the number of people $N$ , occupies one line.

Followed by $N$ number, which means$N$ person's weight$W_1-W_n$。

Output format

For each sample output one line, an integer represents the absolute value of the difference between the weights of the two sides.

Sample

Sample input

1
3
55 50 100

Sample output

5

Data range and tips

对于 100% 的数据，有 $1<=T<=50，2<=N<=20，30<=W_i<=120$。

Ideas

There are only two situations for each person on the left team and the right team.
Depth-first search O(2 ^N );

Code

#include<iostream>
#include<cmath>
using namespace std;
int n,sum,w[21],ans;
void DFS(int dep,int s,int rs)//dep当前人，s一边体重和，rs一边人数和 
{
    
    
	if(dep>n)
	{
    
    
		ans=min(ans,abs(sum-(s<<1)));
		return;
	}
	if(rs<(n+1)>>1)//可以加人 
		DFS(dep+1,s+w[dep],rs+1);
	if(dep-rs<=(n+1)>>1)
		DFS(dep+1,s,rs);
	return;
}
int main()
{
    
    
	int T,i;
	ios::sync_with_stdio(false);
	for(cin>>T;T--;)
	{
    
    
		for(cin>>n,sum=0,ans=100000000,i=1;i<=n;i++)
			cin>>w[i],sum+=w[i];
		DFS(1,0,0),cout<<ans<<endl;
	}
	return 0;
}