题意：给定一个的矩阵, 你有k次机会, 每次机会可以选择某一行或者某一列, 然后使该行或该列的值全部变为0, 并且会得到收益:该行或该列的数字和

Face

数据范围 一开始没看到,,,服了

Strategy: 其实一开始没啥思路, 后来看到了数据范围后想到先二进制枚举选的行数,然后列数贪心选最多

#include <bits/stdc++.h>
#include <bits/extc++.h>

using namespace std;
#define _rep(n, a, b) for (ll n = (a); n <= (b); ++n)
#define _rev(n, a, b) for (ll n = (a); n >= (b); --n)
#define _for(n, a, b) for (ll n = (a); n < (b); ++n)
#define _rof(n, a, b) for (ll n = (a); n > (b); --n)
#define oo 0x3f3f3f3f3f3f
#define ll long long
#define db double
#define eps 1e-8
#define bin(x) cout << bitset<10>(x) << endl;
#define what_is(x) cerr << #x << " is " << x << endl
#define met(a, b) memset(a, b, sizeof(a))
#define all(x) x.begin(), x.end()
#define pii pair<ll, ll>
#define pdd pair<db, db>
const ll maxn = 1e5 + 10;
const ll mod = 1e9 + 7;
ll mat[20][20], n, m, k, sum_ln[20];
signed main(){
    cin >> n >> m >> k;
    _rep(i, 1, n){
        _rep(j, 1, m){
            cin >> mat[i][j];
            sum_ln[i] += mat[i][j];
        }
    }
    ll res = 0;
    _rep(i, 0, (1 << n)-1){
        ll cur = i, used_line = __builtin_popcount(cur);
        if(used_line >k)continue;
        ll val = 0;
        vector<ll> line(20);
        _for(j, 0, n){
            if((cur>>j) &1){
                val += sum_ln[j+1];
                line[j+1] = 1;
                used_line++;
            }
        }
        vector<ll> a(m+1);
        _rep(j, 1, m){
            _rep(k, 1, n){
                if(line[k])continue;
                else a[j] += mat[k][j];
            }
        }
        sort(a.begin() + 1, a.end(), greater<ll>());
        _rep(i, 1, min(m, k - used_line)){
            val += a[i];
        }
        res = max(val, res);
    }
    cout << res << endl;
}