【组队赛三】-D 优先队列 cf446B

DZY Loves Modification
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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Status

Practice

CodeForces 446B
Description
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 Input
Input
2 2 2 2
1 3
2 4
Output
11
Input
2 2 5 2
1 3
2 4
Output
11
Hint
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

<span style="color:#3333ff;">/*
_______________________________________________________________________________________

       author    :   Grant yuan
       time      :   2014.7.21
       algorithm :   priority_queue
       explain   :   首先对行和列在1到k的范围内分别求解，然后求出一个满足i和k-i的最大值，
                     注意终于的结果中减去反复计算的，还有会測试大于int类型的数据。 
—————————————————————————————————————————————————————————————————————————————————————————
*/
    
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<functional>
using namespace std;
#define INF 999999999;

priority_queue<__int64>q1;
priority_queue<__int64>q2;
__int64 m,n,k,p,h;
__int64 a[1003][1003];
__int64 s1[1003],s2[1003];
__int64 sum1[1000003],sum2[1000003];
__int64 res,ans,hh;

int main()
{
    cin>>n>>m>>k>>p;
    ans=INF;
    ans=-ans*ans;
    memset(s1,0,sizeof(s1));
    memset(s2,0,sizeof(s2));
    memset(sum1,0,sizeof(sum1));
    memset(sum2,0,sizeof(sum2));
    while(!q1.empty())
      q1.pop();
    while(!q2.empty())
       q2.pop();

    for(int i=0;i<n;i++)
      for(int j=0;j<m;j++){
        scanf("%I64d",&a[i][j]);
        s1[i]+=a[i][j];
        s2[j]+=a[i][j];}

    for(int i=0;i<n;i++)
      q1.push(s1[i]);

     for(int i=0;i<m;i++)
       q2.push(s2[i]);

    for(int i=1;i<=k;i++)
    {
        h=q1.top();q1.pop();
        if(i==1)sum1[i]=h;
        else sum1[i]=sum1[i-1]+h;
        h-=m*p;
        q1.push(h);
    }
     for(int i=1;i<=k;i++)
    {
        h=q2.top();q2.pop();
        if(i==1)sum2[i]=h;
        else sum2[i]=sum2[i-1]+h;
        h-=n*p;
        q2.push(h);
    }

    for(int i=0;i<=k;i++)
        {
            res=sum1[i]+sum2[k-i];
            ans=max(ans,res-i*p*(k-i));
        }

    cout<<ans<<endl;
    return 0;

}
</span>