题目链接

题意：





题解：









AC代码

/*
    Author : zzugzx
    Lang : C++
    Blog : blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;

#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define endl '\n'
#define SZ(x) (int)x.size()
#define mem(a, b) memset(a, b, sizeof(a))

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod = 1e9 + 7;
//const int mod = 998244353;

const double eps = 1e-6;
const double pi = acos(-1.0);
const int maxn = 1e6 + 10;
const int N = 3e2 + 5;
const ll inf = 0x3f3f3f3f;
const int dir[][2]={{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};

int a[N][N], dp[N][N];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
//  freopen("in.txt", "r", stdin);
//  freopen("out.txt", "w", stdout);
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            cin >> a[i][j];
    for (int i = 1; i <= n; i++)
        sort(a[i] + 1, a[i] + 1 + m);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            a[i][j] += a[i][j - 1];
    mem(dp, inf);
    dp[0][0] = 0;
    for (int i = 1; i <= n; i++)
        for (int j = i; j <= min(n, i * m); j++)
            for (int k = i - 1; k <= min(n, min(j, (i - 1) * m)); k++)
                dp[i][j] = min(dp[i][j], dp[i - 1][k] + a[i][j - k] + (j - k) *(j - k));
    cout << dp[n][n];
    return 0;
}