思路：和背包长得像，那么我们就往背包上想，dp[i][j]表示前i天我们买了j个糖果的最小花费，dp[i][j]可由dp[i-1][k]转移得到，那么怎么转移？j的范围要<=min(n,i*m)k就最小从i-1个糖果开始（再小就不符合题意）最大到min(j,min(n,(i-1)m))当第i-1天的k个糖果取到第i天的j个糖果就需要加上此时在第i天买糖果的花费就是+前(j-k)个糖果的价格（我们用前缀和去维护），再加上（j-k）的平方。得到dp方程为dp[i][j]=min(dp[i][j],dp[i-1][k]+a[i][j-k]+(j-k)(j-k));

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include<iostream>
#include<vector>
#include<queue>
//#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define SIS std::ios::sync_with_stdio(false)
#define space putchar(' ')
#define enter putchar('\n')
#define lson root<<1
#define rson root<<1|1
typedef pair<int,int> PII;
const int mod=1e9+7;
const int N=2e6+10;
const int M=2e3+10;
const int inf=0x7f7f7f7f;
const int maxx=2e5+7;ll gcd(ll a,ll b)
{return b==0?a:gcd(b,a%b);
}ll lcm(ll a,ll b)
{return a*(b/gcd(a,b));
}template <class T>
void read(T &x)
{char c;bool op = 0;while(c = getchar(), c < '0' || c > '9')if(c == '-')op = 1;x = c - '0';while(c = getchar(), c >= '0' && c <= '9')x = x * 10 + c - '0';if(op)x = -x;
}
template <class T>
void write(T x)
{if(x < 0)x = -x, putchar('-');if(x >= 10)write(x / 10);putchar('0' + x % 10);
}
ll qsm(int a,int b,int p)
{ll res=1%p;while(b){if(b&1) res=res*a%p;a=1ll*a*a%p;b>>=1;}return res;
}int n,m;
int dp[310][310],a[310][310];
int main()
{cin>>n>>m;memset(dp,inf,sizeof dp);for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){cin>>a[i][j];}sort(a[i]+1,a[i]+1+m);for(int j=1;j<=m;j++)a[i][j]+=a[i][j-1];}// for(int i=1;i<=n;i++)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]<<endl;return 0;
}
/*
1
3 2
1 1 2 3 4
1 0 1 2 3
2 1 2 3 4
*/