Codeforces Round #FF (Div. 1) B. DZY Loves Modification

매에서 몇 번을 찾았는지 만약에 행에서 i번을 찾았을 때 열은 k-i번을 얻었다. 행렬이 단독으로 욕심을 부리고 덧붙인다고 가정하면 i*(k-i)의 교점이 많다. dpr[i]+dpc[k-i]-i*(k-i)*p
매거 i 최대치...
B. DZY Loves Modification
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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:
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. 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.
Input
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
Output a single integer — the maximum possible total pleasure value DZY could get.
Sample test(s)
input

2 2 2 2
1 3
2 4

output

11

input

2 2 5 2
1 3
2 4

output

11

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

1 1
0 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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

typedef long long int LL;

const int maxn=1100;

LL dpr[maxn*maxn],dpc[maxn*maxn];
LL col[maxn],row[maxn];
LL a[maxn][maxn];
priority_queue<LL> qc,qr;
int n,m,k; LL p;

int main()
{
    scanf("%d%d%d%I64d",&n,&m,&k,&p);
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            scanf("%I64d",&a[i][j]);
            col[j]+=a[i][j];
            row[i]+=a[i][j];
        }
    }
    for(int i=0;i<n;i++)
        qr.push(row[i]);
    for(int i=0;i<m;i++)
        qc.push(col[i]);
    dpr[0]=dpc[0]=0;
    for(int i=1;i<=k;i++)
    {
        LL cc=qc.top(); qc.pop();
        LL rr=qr.top(); qr.pop();
        dpr[i]=dpr[i-1]+rr;
        dpc[i]=dpc[i-1]+cc;
        qc.push(cc-p*n);
        qr.push(rr-p*m);
    }
    LL ans=dpr[0]+dpc[k];
    for(int i=1;i<=k;i++)
        ans=max(ans,dpr[i]+dpc[k-i]-1LL*i*(k-i)*p);
    printf("%I64d",ans);
    return 0;
}