首先要知道选择行列操作时顺序是无关的

用两个数组row[i],col[j]分别表示仅选择i行能得到的最大值和仅选择j列能得到的最大值

这个用优先队列维护，没选择一行(列)后将这行(列)的和减去对应的np (mp)又一次增加队列

枚举选择行的次数为i,那么选择列的次数为k - i次，ans = row[i] + col[k - i] - (k - i) * i * p;

既然顺序无关，能够看做先选择完i次行，那么每次选择一列时都要减去i * p，选择k - i次列，即减去(k - i) * i * p

//#pragma comment(linker, "/STACK:102400000,102400000")

//HEAD

#include <cstdio>

#include <cstring>

#include <vector>

#include <iostream>

#include <algorithm>

#include <queue>

#include <string>

#include <set>

#include <stack>

#include <map>

#include <cmath>

#include <cstdlib>

using namespace std;

//LOOP

#define FE(i, a, b) for(int i = (a); i <= (b); ++i)

#define FED(i, b, a) for(int i = (b); i>= (a); --i)

#define REP(i, N) for(int i = 0; i < (N); ++i)

#define CLR(A,value) memset(A,value,sizeof(A))

//STL

#define PB push_back

//INPUT

#define RI(n) scanf("%d", &n)

#define RII(n, m) scanf("%d%d", &n, &m)

#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)

#define RS(s) scanf("%s", s)

#define FF(i, a, b) for(int i = (a); i < (b); ++i)

#define FD(i, b, a) for(int i = (b) - 1; i >= (a); --i)

#define CPY(a, b) memcpy(a, b, sizeof(a))

#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)

#define EQ(a, b) (fabs((a) - (b)) <= 1e-10)

#define ALL(c) (c).begin(), (c).end()

#define SZ(V) (int)V.size()

#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)

#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)

#define WI(n) printf("%d\n", n)

#define WS(s) printf("%s\n", s)

#define sqr(x) x * x

typedef vector <int> VI;

typedef unsigned long long ULL;

typedef long long LL;

const int INF = 0x3f3f3f3f;

const int maxn = 1010;

const double eps = 1e-10;

const LL MOD = 1e9 + 7;

int ipt[maxn][maxn];

LL row[maxn * maxn], col[maxn * maxn];

LL rtol[maxn], ctol[maxn];

int main()

{

    int n, m, k, p;

    while (~RIV(n, m, k, p))

    {

        priority_queue<LL> r, c;

        int radd = 0, cadd = 0;

        CLR(rtol, 0), CLR(ctol, 0);

        FE(i, 1, n)

            FE(j, 1, m)

            {

                RI(ipt[i][j]);

                rtol[i] += ipt[i][j];

                ctol[j] += ipt[i][j];

            }

        FE(i, 1, n)

            r.push(rtol[i]);

        FE(j, 1, m)

            c.push(ctol[j]);

        row[0] = 0, col[0] = 0;

        FE(i, 1, k)

        {

            LL x = r.top(), y = c.top();

             r.pop(), c.pop();

            r.push(x - m * p);

            c.push(y - n * p);

            row[i] = row[i - 1] + x;

            col[i] = col[i - 1] + y;

        }

//        FE(i, 0, k)

//            cout << row[i] + col[k - i] <<endl;

        LL ans = -1e18;

        FE(i, 0, k)

            ans = max(ans, row[i] + col[k - i] - (LL)i * (k - i)* p);

        cout << ans << endl;

    }

    return 0;

}

/*

2 2 4 2

1 2 3 10

2 3 5 2

2 2 2

2 2 2

*/