2019牛客多校2 F Partition problem(dfs)

题意：

n<=28个人，分成人数相同的两组，给你2*n*2*n的矩阵，如果(i,j)在不同的组里，竞争力增加v[i][j]，问你怎么分配竞争力最

4s

思路：

枚举C(28,14)的状态，更新答案即可

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
//#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-;
const int mod = 1e9+;
const int maxn = 1e6+;
const int maxm = 2e6+;
const int inf = 0x3f3f3f3f;
//const db pi = acos(-1.0);

int n;
ll v[][];
ll ans;
ll tmp;
int cnt0,cnt1;
int v0[maxn],v1[maxn];
int m;
void dfs(int x, int y){//p / num of need 1
    ll now=;
    if(x==m+){ans=max(ans,tmp);return;}
    if(m-x>=y){
        v0[++cnt0]=x;
        for(int i = ; i <= cnt1; i++)now+=v[v1[i]][x];
        tmp+=now;
        dfs(x+,y);
        tmp-=now;
        cnt0--;
    }
    if(y>){
        now=;
        v1[++cnt1]=x;
        for(int i = ; i <= cnt0; i++)now+=v[v0[i]][x];
        tmp+=now;
        dfs(x+,y-);
        tmp-=now;
        cnt1--;
    }
    return;
}
int main(){
    scanf("%d", &n);
    m = *n;
    ans=tmp=;
    cnt0=cnt1=;
    for(int i = ; i <= m; i++){
        for(int j = ; j <= m; j++){
            scanf("%d", &v[i][j]);
        }
    }
    dfs(,n);
    printf("%lld",ans);
    return ;
}
/*
2
0 1 1 1
1 0 1 1
1 1 0 1
1 1 1 0
 */