题目意思：

Solution

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define mk(__x__,__y__) make_pair(__x__,__y__)
#define ms(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x, int op = 10) { if (!x) { putchar('0'); if (op)    putchar(op); return; }    char F[40]; ll tmp = x > 0 ? x : -x;    if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]);    if (op)    putchar(op); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
const int dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;

const int N = 300 + 7;
ll a[N][N];
ll dp[N][N];

int main() {
    int n = read(), m = read();
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j)
            a[i][j] = read();

    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];

    ms(dp, 0x3f);
    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));

    print(dp[n][n]);
    return 0;
}