As we know, DZY loves playing games. One day DZY decided to play with a n × m matrix. To be more precise, he decided to modify the matrix with exactly k operations.

Each modification is one of the following:

1. Pick some row of the matrix and decrease each element of the row by p. This operation brings to DZY the value of pleasure equal to the sum of elements of the row before the decreasing.
2. Pick some column of the matrix and decrease each element of the column by p. This operation brings to DZY the value of pleasure equal to the sum of elements of the column before the decreasing.

DZY wants to know: what is the largest total value of pleasure he could get after performing exactly k modifications? Please, help him to calculate this value.

The first line contains four space-separated integers n, m, k and p (1 ≤ n, m ≤ 103; 1 ≤ k ≤ 106; 1 ≤ p ≤ 100).

Then n lines follow. Each of them contains m integers representing aij (1 ≤ aij ≤ 103) — the elements of the current row of the matrix.

Output a single integer — the maximum possible total pleasure value DZY could get.

For the first sample test, we can modify: column 2, row 2. After that the matrix becomes:

1 10 0

For the second sample test, we can modify: column 2, row 2, row 1, column 1, column 2. After that the matrix becomes:

-3 -3-2 -2

题解：

首先上一个错误的想法，分别统计出每一行和每一列的和，然后用优先队列维护，每次取出最大值。

将所对应的行或列的每一个值-P，再更新一下答案。

这种做法为什么是错的呢，因为假设如果有一行和一列的和是相等的，那么我们并不知道要先走行还是先走列。

再来看正确的做法：

首先我们设最终选了 行 i 次，则列选了 k-i 次

那么假设我们先全部选行，然后选列，则每次选列时，要-= i*p

这样最后是 -= i*(k-i)*p

也就是所有行对列的影响

那我们先把这个 i*(k-i)*p 提出来，那么选行和选列就互不影响

就可以分别考虑行和列

对于只取行的情况：

预处理出选0次 1次······k次行的最大值 H[i]

即优先队列跑一次即可

同理对列处理, 得到 L[i] 表示取i次列, 0次行的最大值

然后 ans = max( H[i]+L[k-i] - i*(k-i)*p )

注意结果可能是很小的负数,ans = -inf ,inf要足够大

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+7;
long long c[maxn],r[maxn],ll[maxn],rr[maxn];
long long a[1005][1005];
int n,m,k,p;
int main()
{scanf("%d%d%d%d",&n,&m,&k,&p);for(int i=1; i<=n; i++){for(int j=1; j<=m; j++){scanf("%I64d",&a[i][j]);ll[i]+=a[i][j];rr[j]+=a[i][j];}}priority_queue<long long>Q;for(int i=1; i<=n; i++)Q.push(ll[i]);for(int i=1; i<=k; i++){int now = Q.top();Q.pop();c[i]=c[i-1]+now;now-=m*p;Q.push(now);}while(!Q.empty())Q.pop();for(int i=1; i<=m; i++)Q.push(rr[i]);for(int i=1; i<=k; i++){int now=Q.top();Q.pop();r[i]=r[i-1]+now;now-=n*p;Q.push(now);}long long ans = -1LL<<60;for(int i=0; i<=k; i++)ans=max(ans,c[i]+r[k-i]-1ll*i*(k-i)*p);cout<<ans<<endl;
}