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)的和减去相应的np (mp)重新加入队列

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，可以看做先选择完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
*/