[C++知识库]矩阵消除游戏

// Problem: 矩阵消除游戏
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/20960/1045
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
// 2022-03-01 19:19:12
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a%mod;
		a = a * a %mod;
		b >>= 1;
	}
	return ans;
}
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template<typename T> void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
	return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
	return (a + b) %mod;
}
inline ll inv (ll a) {
	return qmi(a, mod - 2);
}
#define int long long
int n, m, k;
int a[20][20], b[20][20];
int h[29], l[20];
void solve() {
	cin >> n >> m >> k;
	int sum = 0;
	for (int i = 0;i < n; i ++) {
		for (int j = 0; j < m; j ++)
			{
				cin >> a[i][j];
				h[i] +=a[i][j];
            sum += a[i][j];
			}
	}
	if (k >= min(n, m)) {
        cout << sum << endl;
        return;
    }
	int ans = 0;
	for (int i = 0; i < (1 << n); i ++) {
		int mx = 0, cnt = 0;
		memcpy(b, a, sizeof b);
		for (int j = 0; j < n; j ++)
			if (i >> j & 1) {
				mx += h[j];
				cnt ++;
				for (int kk = 0; kk < m; kk ++)
					b[j][kk] = 0;
			}
		if (cnt > k) continue;
		memset(l, 0, sizeof l);
		for (int j = 0; j < m; j ++) {
			for (int k = 0; k < n; k ++)
				l[j] += b[k][j];
		}
		sort(l, l + m, greater<int>());
		for (int kk = 0; kk < k - cnt; kk ++)
			mx += l[kk];
		ans = max (ans, mx);
	}
	cout << ans << endl;
}
signed main () {
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

由于最终一定是选择了一部分列和一部分行。因此我们可以先确定行的选法然后从大到小选则列。 由于n<=15所以可以用二进制枚举