CF446B DZY Loves Modification 优先队列

CF446B DZY Loves Modification 优先队列

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 a_ij (1 ≤ a_ij ≤ 10³) — the elements of the current row of the matrix.

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

2 2 2 2
1 3
2 4

11

2 2 5 2
1 3
2 4

11

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
貌似行和列不太好处理？
假设先对行进行处理了 i 次，那么列自然就是 k-i 次处理；
对行操作结束后： sum-=m*p*i;
此时对列的就是 -= (k-i)*p*i;
那么我们枚举行和列的操作次数即可；

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