1
  2
  3
  4
  5
  6
  7
  8
  9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
 51
 52
 53
 54
 55
 56
 57
 58
 59
 60
 61
 62
 63
 64
 65
 66
 67
 68
 69
 70
 71
 72
 73
 74
 75
 76
 77
 78
 79
 80
 81
 82
 83
 84
 85
 86
 87
 88
 89
 90
 91
 92
 93
 94
 95
 96
 97
 98
 99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
#include<bits/stdc++.h>
using namespace std;

#define rep(i, a, b) for (int i = (a); i < (b); i++)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
#define ST first
#define ND second
#define PB push_back
using ll = long long;
using ld = long double;
using pi = pair<int,int>;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;


#ifdef LOCAL
#define DTP(x, y)                                    \
    auto operator<<(auto &o, auto a)->decltype(y, o) \
    {                                                \
        o << "(";                                    \
        x;                                           \
        return o << ")";                             \
    }
DTP(o << a.first << ", " << a.second, a.second);
DTP(for (auto i : a) o << i << ", ", all(a));
void dump(auto... x) { ((cerr << x << ", "), ...) << '\n'; }
#define debug(x...) cerr << setw(4) << __LINE__ << ":[" #x "]: ", dump(x)
#else
#define debug(...) 0
#endif

bool cmp(vll &A, vll &B){
    return (A[0]>B[0]);
}

void Solve()
{
    ll n, m, k;
    cin >> n >> m >> k;
    vector<vector<ll>> A;
    vector<vector<ll>> R;
    rep(j, 0, n){
        vector<ll> T(m);
        bool rev=true;
        rep(i, 0, m){
            cin >> T[i];
            if(i && T[i]>T[i-1])
                rev=false;
        }
        reverse(all(T));
        T.PB(0);
        if(rev)
            R.PB(T);
        for(int i=m-1; i>=0; --i)
            T[i]+=T[i+1];
        if(!rev)
            A.PB(T);
    }
    vector<ll> Rev_ans(sz(R)*m+1);
    vi I(sz(R), m-1);
    priority_queue<pair<ll,int>> Q;
    rep(i, 0, sz(R)){
        Q.push(pair<ll,int>(R[i][m-1], i));
    }
    rep(i, 1, sz(R)*m+1){
        pair<ll,int> x=Q.top();
        Q.pop();
        Rev_ans[i]=Rev_ans[i-1]+x.ST;
        if(I[x.ND]>0){
            --I[x.ND];
            Q.push(pair<ll,int>(R[x.ND][I[x.ND]], x.ND));
        }    
    }
    n=sz(A);
    sort(all(A), cmp);
    vll Ans(n*m+1);
    if(n>0){
        vll Pref(n);
        Pref[0]=A[0][0];
        rep(i, 1, n)
            Pref[i]=Pref[i-1]+A[i][0];
        Ans[n*m]=Pref[n-1];
        for(ll i=0; i<m; ++i){ // Bierzemy pierwsze do j-1 włącznie, z reszty 1 wierzch dł. i
            ll ma=0;
            for(ll j=n-1; j>=0; --j){
                ma=max(ma, A[j][m-i]);
                if(j==0){
                    Ans[i]=ma;
                }else{
                    Ans[(j)*m+i]=Pref[j-1]+ma;
                }
            }
        }
        for(ll i=1; i<m; ++i){ // Z pierwszych do j-tego włącznie z jednego bierzemy tylko m-i górnych
            ll mi=1000'000'000'000'000'000;
            for(ll j=0; j<n; ++j){
                mi=min(mi, A[j][0]-A[j][i]);
                Ans[(j+1)*m-i]=max(Ans[(j+1)*m-i], Pref[j]-mi);
            }
        }
    }
    ll res=0;
    rep(i, 0, sz(Ans)){
        if(k-i>=0 && k-i<sz(Rev_ans) && res<Ans[i]+Rev_ans[k-i])
            res=Ans[i]+Rev_ans[k-i];
    }
    cout << res << "\n";
    return;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int t=1;
    // cin >> t;
    while(t--)
    {
        Solve();
    }
    return 0;
}