19牛客多校（第二场）F Partition problem

链接：https://ac.nowcoder.com/acm/contest/882/F
来源：牛客网

时间限制：C/C++ 4秒，其他语言8秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

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 ∑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.

示例1

输入

复制

1
0 3
3 0

输出

复制

这道题快把我做she了。。。。C(14,28)的值4e7左右，他的时限是在4秒左右，直接暴力跑所有情况是能跑完的。

但是如果每种情况都要n*n的暴力求总竞争值，那么会超时：复杂度多了。这种情况下我们可以看出来，在递归过程中下一次状态的竞争值和上一个状态的值差的就是加上某个点之后的变化。。所以我们可以通过一些操作来减少一些复杂度的。~就酱~~

PS:服了~~沙雕代码编辑器快吐了~


	
	
		
		
			
			
				#include<bits/stdc++.h>
			

			
				usingnamespacestd;
			

			
				intar[17], br[17];
			

			
				intmp[30][30], n; longlongans = 0;
			

			
				intacnt = 0, bcnt = 0;
			

			
				voidgo(intpos, longlongres) {
			

			
				    if(pos >= 2 * n) {
			

			
				        ans = max(res, ans);
			

			
				        return;
			

			
				    }
			

			
				    if(acnt < n) {
			

			
				        ar[++acnt] = pos;
			

			
				        longlongtmp = 0;
			

			
				        for(inti = 1; i <= bcnt; i++) {
			

			
				            tmp += mp[pos][br[i]];
			

			
				        }
			

			
				        go(pos + 1, res + tmp);
			

			
				        --acnt;
			

			
				    }
			

			
				    if(bcnt < n) {
			

			
				        br[++bcnt] = pos;
			

			
				        longlongtmp = 0;
			

			
				        for(inti = 1; i <= acnt; i++) {
			

			
				            tmp += mp[pos][ar[i]];
			

			
				        }
			

			
				        go(pos + 1, res + tmp);
			

			
				        --bcnt;
			

			
				    }
			

			
				}
			

			
				intmain() {
			

			
				    scanf("%d", &n);
			

			
				    for(inti = 0; i < 2*n; i++) {
			

			
				        for(intj = 0; j < 2*n; j++) {
			

			
				            scanf("%d", &mp[i][j]);
			

			
				        }
			

			
				    }
			

			
				    go(0, 0LL);
			

			
				    printf("%lld\n", ans);
			

			
				    return0;
			

			
				}
			

			
				/*
			

			
				2
			

			
				0 1 2 3
			

			
				1 0 4 5
			

			
				2 4 0 6
			

			
				3 5 6 0
			

			
				 
			

			
				*/