思路:
先枚举行,在确定行的基础上去贪心列。
用01字符串去枚举,这个还是头会见,毕竟菜的扣脚。
思路不是很难懂。
#include<iostream>
#include<string>
#include<math.h>
#include<algorithm>
#include<vector>
//#include<bits/stdc++.h>
typedef long long ll;
ll gcd(ll a, ll b)
{
return b ? gcd(b, a % b) : a;
}
ll lcm(ll a, ll b) {
return a * b / (gcd(a, b));
}
#define PII pair<int,int>
using namespace std;
const int N = 2e6 + 10, mod = 1e9 + 7;
int qmi(int a, int k, int p) //快速幂模板
{
int res = 1;
while (k)
{
if (k & 1) res = (ll)res * a % p;
k >>= 1;
a = (ll)a * a % p;
}
return res;
}
///////////////////////////////////////////////////////////
int n, m, k,an[25][25];
long long he[25], sumlie[25],ans = 0;
bool b[25];
int solve(int x)
{
memset(b, 0, sizeof(b));
int answer = 0, i = 1;
while (x)
{
if (x & 1) //末尾为1
{
answer++;
b[i] = 1;
}
x >>= 1;
i++;
}
return answer;
}
int main()
{
cin >> n >> m >> k;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
{
cin >> an[i][j];
he[i] += an[i][j];
}
if (k > n || k > m) k = max(n, m);
int tmp = (1 << n) - 1;
for (int t = 0; t <= tmp; t++)
{
int hang = solve(t);
int lie = k - hang;
if (lie > m || lie < 0) continue;
long long sum = 0;
for (int i = 1; i <= n; i++)
if (b[i]) sum += he[i];
memset(sumlie, 0, sizeof(sumlie));
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
if (b[i]==0) sumlie[j] += an[i][j];
sort(sumlie + 1, sumlie + 1 + m);
for (int i = 1, j = m; i <= lie; i++, j--) sum += sumlie[j]; //直接贪心处理
ans = max(ans, sum);
}
cout << ans << endl;
return 0;
}

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