2019牛客多校训练营第二场 F. Partition problem

题目链接

题目描述

Given $2N$ people, you need to assign each of them into either red team or white team such that each

team consists of exactly $N$ people and the total competitive value is maximized.

Total competitive value is the summation of competitive value of each pair of people in different team.

The equivalent equation is ${\sum }_{i=1}^{2N}{\sum }_{j=i+1}^{2N}{v}_{ij}$(if i-th person is not in the same team as j-th person else 0)

输入描述

The first line of input contains an integers $N$.

Following $2N$ lines each contains $2N$ space-separated integers $v_{ij}$

• $1\le N\le 14$
• $0\le v_{ij}\le 10^9$
• $v_{ij}=v_{ji}$

输出描述

Output one line containing an integer representing the maximum possible total competitive value.

示例1

输入
1
0 3
3 0
输出
3

题目大意

一只球队总共有$2\times n$个队员，每两个队员之间都有一个竞争值，将$2\times n$个队员分成两组，使得不同组队员之间竞争值和最大。

解题思路

拿到这道题的时候就想到了全排列，也就是暴力搜索，$N$最大为14，那么时间复杂度就是$C(28,14)$，值在$4e7$左右，那么接下来就是计算所有情况下的竞争值总和，如果直接用双层循环来计算的话，时间复杂度$O(n^2)$,即$196$，直接会炸掉，所以不能在得到分组结果之后才进行计算竞争值之和，这样会付出更多的时间代价。而正解应该是一边进行分组，一边进行计算。

AC代码

#include
#define _MATH_DEFINES_DEFINED
const int Max_N=1e5+10;
const int mod=998244353;
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
int team[30][30];
int team1[30];
int team2[30];
ll res=0;
int n;
void dfs(int num1,int num2,int pos,ll maxres)
{
	if(pos==2*n)
	{
		res=max(res,maxres);
	}
	if(num1<n)
	{
		team1[num1]=pos;
		ll v=maxres;
		for(int i=0;i<num2;i++)
			v+=team[pos][team2[i]];
		dfs(num1+1,num2,pos+1,v);
	}
	if(num2<n)
	{
		team2[num2]=pos;
		ll v=maxres;
		for(int i=0;i<num1;i++)
			v+=team[pos][team1[i]];
		dfs(num1,num2+1,pos+1,v);
	}
	return ;
}
int main()
{
	cin>>n;
	for(int i=0;i<2*n;i++)
	{
		for(int j=0;j<2*n;j++)
		{
			cin>>team[i][j];
		}
	}
	dfs(0,0,0,0);
	cout<<res<<endl;
}

总结

暴力的题目不能过分依赖$STL$的函数，手写$dfs$并加入优化的过程会得到意想不到的结果。