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 ∑2Ni=1∑2Nj=i+1(vij if i-th person is not in the same team as j-th person else 0)∑i=12N∑j=i+12N(vij 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 vijvij is the j-th value of the i-th line which indicates the competitive value of person i and person j.
* 1≤N≤141≤N≤14 * 0≤vij≤1090≤vij≤109 * vij=vjivij=vji

输出描述: 

Output one line containing an integer representing the maximum possible total competitive value.
题意：给出2*n个人，以及给出他们之间的关系数值，现在要求将这些人分成两拨，一拨有n个人，然后求
这个最大
我们考虑对于集合A,B是不是等价于集合B,A，所以我们一开始将1放入A中，其余都放入B中，然后按照顺序从前向后枚举编号，将之从B中弹出，加入到A中，一个元素从B中弹出就需要减去与原A中元素的连边，加上与B中其他元素的连边即可。
剪枝：可行性剪枝，如果将剩下的所有数字都加入到A中都无法凑够n个人说明不可行，剪掉。
时间复杂度为,这时间复杂度看上去不对劲，事实上加上剪枝可以3秒跑完emmm
方案数为组合数，算答案为2*n。
下面贴代码： 

	#include<bits/stdc++.h>





	usingnamespacestd;





	#define ll long long





	#define INF 0x3fffffff





	#define maxn 100005





	#define mod 1000000000





	intn;





	ll a[30][30],ans;





	//参数依次为:msk二进制下为1的位置为A中的元素,num为A中的元素个数,





	//pre表示区间[0,pre]已被使用,ans2为当前答案





	voidac(intmsk,intnum,intpre,ll ans2){





	    if(num<<1==n){





	        ans=max(ans,ans2);





	        return;





	    }





	    if(((pre+1-num)<<1)>n)





	        return;





	    for(intnxt=pre+1;nxt<n;++nxt){





	        ll t=ans2;





	        for(inti=0;i<n;++i){





	            if((msk>>i)&1)





	                t-=a[nxt][i];//减掉与A中的连边;





	            else





	                t+=a[nxt][i];//加上与B中的连边;





	        }





	        ac(msk|(1<<nxt),num+1,nxt,t);





	    }





	}





	intmain(){





	    scanf("%d",&n);





	    n<<=1;





	    for(inti=0;i<n;++i)





	        for(intj=0;j<n;++j)





	            cin>>a[i][j];





	    ll ans2=0;





	    for(inti=0;i<n;++i)





	        ans2+=a[0][i];





	    ac(1,1,0,ans2);





	    printf("%lld\n",ans);





	}