D. DZY Loves Modification

D. 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:

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.

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

文章大意是给你一个n * m的矩阵,你能够进行k次操作.

操作1:把一行的每一个元素都减去p,你能够获得该行全部元素和(操作之前)的pleasure

操作2:把一列的每一个元素都减去p,你能够获得该列全部元素和(操作之前)的pleasure

问你最大能够获得的pleasure是多少

思路:枚举操作1所进行的次数,每次取能获得最大的行,用优先队列维护,要注意预处理,否则会超时

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
typedef long long LL;
using namespace std;
int A[1005][1005];
LL R[1005],C[1005];
LL s1[1000005],s2[1000005];
void get_C(int n,int i,int (*A) [1005])
{
    LL s=0;
    for(int j=1;j<=n;j++)
        s+=A[j][i];
    C[i]=s;
}
void get_R(int m,int i,int (*A)[1005])
{
    LL s=0;
    for(int j=1;j<=m;j++)
        s+=A[i][j];
    R[i]=s;
}
int main()
{
    int n,m,k,p;
    while(scanf("%d%d%d%d",&n,&m,&k,&p)==4)
    {
       LL s=0;
       s1[0]=s2[0]=0;
       for(int i=1;i<=n;i++)
       for(int j=1;j<=m;j++)
           scanf("%d",&A[i][j]);
        LL sum=-1000000000000009;
        priority_queue<LL>q1;
        priority_queue<LL>q2;
        for(int i=1;i<=n;i++){get_R(m,i,A);q1.push(R[i]);}
        for(int i=1;i<=m;i++){get_C(n,i,A);q2.push(C[i]);}
        for(int j=1;j<=k;j++)
        {
               LL temp=q1.top();
               q1.pop();
               s1[j]=s1[j-1]+temp;
               temp-=p*m;
               q1.push(temp);
        }
        for(int j=1;j<=k;j++)
        {
               LL temp=q2.top();
               q2.pop();
               s2[j]=s2[j-1]+temp;
               temp-=p*n;
               q2.push(temp);
        }
        for(int i=0;i<=k;i++)
        {
           s=s1[i]+s2[k-i];
           sum=max(sum,s-LL(i)*p*(k-i));
        }
       printf("%I64d\n",sum);
    }
    return 0;
}